Suppose the high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. at which time the water is 10 feet above the height of the low tide. Low tides occur 6 hours after high tides. Suppose there are two high tides and two low tides every day and the height of the tide varies sinusoidally.
a) Find a formula for the function y = h(t) that computes the height of the tide above low tide at time t. (In other words, y = 0 corresponds to low tide)
b) What is the tide height at 11:00 am?

Answer:

-26

Step-by-step explanation:

y=mx+c

m= (y2 - y1 ) / (x2-x1)

m= (10-1) / (48 -36)

m = 0.75

10 = 0.75(48) + c

c = 10 - 36

c = -26

Answer:

1)   y = x + 3

2)   31 years old

Step-by-step explanation:

Jack is x years old

Todd is y years old

Todd is 3 years older than his brother Jack => y = x + 3

When Jack is 28 years old, Todd is y = 28 + 3 = 31 years old

Answer:

[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]

And we can find this probability with this difference:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]

And if we use the normal standard distribution or excel we got:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]

Step-by-step explanation:

Let X the random variable that represent the time required to construct and test a particular component of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(60,9.4)[/tex]  

Where [tex]\mu=60[/tex] and [tex]\sigma=9.4[/tex]

We want to find this probability:

[tex]P(52<X<70)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using this formula we got:

[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]

And we can find this probability with this difference:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]

And if we use the normal standard distribution or excel we got:

[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]

Answer:

work is shown and pictured

Answer:

80 grams

Step-by-step explanation:

Weight of 24% solution = 400 grams

Alcohol content= 24%

Amount of alcohol= 400*0.24= 96 grams

Weight of 30% solution with same amount of alcohol:

The difference in weights is the other chemicals evaporated:

Answer:

99.27%

Step-by-step explanation:

We have to:

sample size (n) = 100

p = 0.8

1 egg = 3 * 0.01 = 0.03

eggs = 100 - 30 broken eggs = 70

Let's find the probability of winning more than $ 1.30 today:

p (x> 1.30) = p (z, (x -m) / sd)

average earnings:

per day would be:

m1 = 100 / 1.5 * 0.8 * 0.03

m1 = 1.6

sd1 = (m1 * (1 - p)) ^ (1/2) = (1.6 * (1 - 0.8)) ^ (1/2)

sd1 = 0.566

earnings for broken eggs:

m2 = 70 * 0.8 * 0.03

m2 = 1.68

sd2 = (m2 * (1 - p)) ^ (1/2) = (1.68 * (1 - 0.8)) ^ (1/2)

sd2 = 0.58

Now the total profit would be:

m = m1 + m2 = 1.6 + 1.68

m = 3.28

sd = sd1 ^ 2 + sd2 ^ 2 = 0.566 ^ 2 + 0.58 ^ 2

sd = 0.81

now yes, replacing:

p (x> 1.30) = p (z> (1.3 - 3.28) /0.81)

p (x> 1.30) = p (z> -2.44)

for this z = 0.0073

Therefore the probability would be:

1 - 0.0073 = 0.9927

That is, we would have a 99.27% probability of achieving the goal.

Answer:

It’s C

Step-by-step explanation:

But thanks for helping me get it wrong toxo360❤️

In the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of: ax² + bx + c = 0.

The number of people attending graduate school at a university may be modeled by the quadratic regression equation y = 10x2 - 40x+8, where x represents the year.

The regression equation, which year is the best prediction for when 2528 people will attend graduate school is;

[tex]\rm y = 10x^2 - 40x+8\\\\2528 = 10x^2 - 40x+8\\\\ 10x^2 - 40x+8\\\\ 10x^2 - 40x+8-2528=0 \\\\ 10x^2 - 40x-2520=0\\\\\text{Divide by 10 both sides}\\\\x^2-4x-252=0\\\\x^2-18x+14x-252=0\\\\x(x-18)+14(x-18)=0\\\\(x-18)(x+14)=0\\\\x-18=0, \ x=18\\\\x+14=0, \ x=-14[/tex]

Hence, the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.

Learn more about quadratic equations here;

https://brainly.com/question/24298861

#SPJ2

Answer:

The miner should take door number 1

Answer:

5y - 4 ≥ 26

5y ≥ 26 + 4

5y ≥ 30

y ≥ 30/5

y ≥ 6

Answer:

[tex]150-2.797\frac{35}{\sqrt{25}}=130.421[/tex]    

[tex]150+2.797\frac{35}{\sqrt{25}}=169.579[/tex]    

Step-by-step explanation:

Information given

[tex]\bar X=150[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean

s=35 represent the sample standard deviation

n=25 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=25-1=24[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex] and the critical value for this case [tex]t_{\alpha/2}=2.797[/tex]

Now we have everything in order to replace into formula (1):

[tex]150-2.797\frac{35}{\sqrt{25}}=130.421[/tex]    

[tex]150+2.797\frac{35}{\sqrt{25}}=169.579[/tex]    

Answer:60°

Step-by-step explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me....

Answer:

A.

60°

Hope its right.

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Since you are given line RT and ND, then all you need to do by ASA is the make sure that the angles are the two endpoints are congruent. Since the problem already gave you R and N, then all thats left is to relate the other two endpoints, namely, T and D

Answer:

2.75 inches

Step-by-step explanation:

City B experienced three times the amount of rainfall than  City A, therefore :

Since both cities received 11 inches of rainfall in total.

We have that:

x+3x=11

4x=11

Divide both sides by 4

x=2.75 Inches

Therefore, we City A received 2.75 inches of rain.

Answer:

See explanation

Step-by-step explanation:

Q1)a

- Denote a random variable ( X ) as the life time of a brand of bulb produced.

- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:

                           X ~ Norm ( 210 , 56^2 )

i) The bulb picked to have a life time of at least 300 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )

          P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )

          P ( X ≥ 300 ) = P ( Z ≥ 1.607 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054  .. Answer

ii) The bulb picked to have a life time of at most 100 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )

          P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )

          P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247  .. Answer

iii) The bulb picked to have a life time of between 150 and 250 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )

          P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )

          P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205  .. Answer

Q1)b

- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:

                    p ( A ) = 0.25

- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:

                    p ( B ) = 0.4

- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.  

- So for independent events the required probability can be stated as:

              p ( A' & B' ) = p ( A' ) * p ( B' )

              p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]

              p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]

              p ( A' & B' ) = 0.45 ... Answer

Q2)a

- A discrete random variable X: defines the probability of getting each number on a biased die.

- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.

             ∑ p ( X = xi ) = 1

             p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )

             1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1

             k = 0.1  ... Answer

- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:

             E ( X ) = ∑ p ( X = xi ) . xi

             E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6

             E ( X ) = 3.5 .. Answer

- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:

             E ( X^2 ) = ∑ p ( X = xi ) . xi^2

             E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36

             E ( X^2 ) = 15.233 .. Answer

- The variance of the discrete random distribution for the variable X can be determined from:

            Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2

            Var ( X ) = 15.2333 - [ 3.5 ] ^2

            Var ( X ) = 2.9833 ... Answer

- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:

           P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )

           P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1

           P ( 1 < X < 5 ) = 0.467  ... Answer

Q2)b

- Two independent events are defined by their probabilities as follows:

           p ( A ) = 0.3  and p ( B ) = 0.5

- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.

- For the two events to occur simultaneously at the same time:

          p ( A & B ) = p ( A )* p ( B )

          p ( A & B ) = 0.3*0.5

          p ( A & B ) = 0.15  ... Answer

- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:

        p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )

        p ( A U B ) = 0.3 + 0.5 - 2*0.15

        p ( A U B ) = 0.5 ... Answer

- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:

       p ( A & B ) = 0.15 ≠ 0

Q2)c

- The letters of the word given are to be arranged in number of different ways as follows:

                                  STATISTICS

- Number of each letters:

        S : 3

        T : 3

        A: 1

        I: 2

        C: 1

- 10 letters can be arranged in 10! ways.

- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:

                        [tex]\frac{10!}{3!3!2!} = \frac{3628800}{72} = 50,400[/tex]

- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.

Answer:

Yes, her reasoning is correct

Step-by-step explanation:

Given the ratio:  109.2 : 6

Written in fractional form, we have:

[tex]109.2: 6=\dfrac{109.2}{6}[/tex]

Now:

[tex]\dfrac{109.2}{6}=\dfrac{109.2}{6}X1, $ Let 1=\dfrac{10}{10}, \\\\=\dfrac{109.2}{6}X\dfrac{10}{10}\\\\=\dfrac{1092}{60}[/tex]

Therefore, the student's reasoning is correct. In fact, as a check:

[tex]\dfrac{109.2}{6}=18.2\\\\\dfrac{1092}{60}=18.2[/tex]

We would obtain the same result in both cases.

Answer:

yes

Step-by-step explanation:

Answer:

-2> 7+35

-2>42

0<21

the answer is 0 is less than 21

Answer:

w = 9, l = 12

Step-by-step explanation:

If w is the width and l is the length, then:

l = 2w − 6

The area of the rectangle is width times length.

A = wl

Substituting and solving:

108 = w (2w − 6)

108 = 2w² − 6w

54 = w² − 3w

0 = w² − 3w − 54

0 = (w + 6) (w − 9)

w = -6 or 9

The width is 9, so the length is 12.

Answer:

These two statements are equivalent because statement A is the contrapositive of statement B.

Step-by-step explanation:

In logic, when we have an "if" statement we can have its contrapositive or its converse.

Given a "if p, then q" (p is the hypothesis and q is the conclusion), the converse is "if q, then p", in other words, we interchange the hypothesis and the conclusion.

Now, given a "if p, then q", the contrapositive is "if not q, then not p". In other words, we take the negation of both the hypothesis and the conclusion and then we interchange them.

Now, let's take a look at our statements:

If a and b are two real numbers, and if ab = 0, then either a = 0 or b = 0.

In this case:

p = a and b are two real numbers and ab=0

q = either a = 0 or b = 0

Now, let's take the negative of p and q:

The negative of p would be: a and b are two real numbers and their product is not zero.

The negative of q would be: neither of the two real numbers is zero

Now, given than the contrapositive is "if not q, then not p" we would have:

If neither of two real numbers is zero, then their product is not zero.

We can see that this last sentence is the contrapositive of the first one and thus:

These two statements are equivalent because statement A is the contrapositive of statement B.

Answer:

Step-by-step explanation:

Total Investment = ​$100,000

The investment was split into three parts (say x, y and z)

Simple Interest = Principal X Rate/100 X Time

The first part of the investment earned​ 8% interest

Interest on the first part = 0.08x

The second part of the investment earned​ 6% interest

Interest on the second  part = 0.06y

The third part of the investment earned​ 9% interest

Interest on the third part = 0.09z

Total interest from the investments = $ 7380.

Therefore:

0.08x+0.06y+0.09z=7380

The interest from the first investment was 2 times the interest from the second.

0.08x=2 X 0.06y

0.08x=0.12y

Substituting we have:

0.12y+0.06y+0.09z=7380

0.18y+0.09z=7380

From 0.08x=0.12y

x=1.5y

Substituting x=1.5y into x+y+z=100000

1.5y+y+z=100000

2.5y+z=100000

z=100000-2.5y

Substituting z=100000-2.5y into 0.18y+0.09z=7380

0.18y+0.09(100000-2.5y)=7380

0.18y+9000-0.225y=7380

-0.045y=-1620

Divide both sides by -0.045

y=36000

Recall: x=1.5y

x=1.5 X 36000

x =54000

x+y+z=100000

54000+36000+z=100000

z=100000-(54000+36000)

z=10,000

Therefore:

Answer:

$1946

Step-by-step explanation:

His average for the first four months

= $1,450.25

So he earned 4*$1,450.25

= $5801 for the first four months.

Then for the year his average is $1,780.75.

So he earned $1,780.75*12 for the year.

=$ 21369

So the amount he earned for the remaining 8 months was 21369-5801

= $15568

The average for the 8 Months= 15568/8

= $1946

Answer:

Step-by-step explanation:

Since we know that for a distribution be a probability density function sum of all the probability events should be equal to 1 and all individual events should have probability between 0 and 1

a. x P(X=x)

0 -----3/8

1 -----1/4

2 -----3/8

P(X=0)+P(X=1)+P(X=2) = 3/8 + 1/4 + 3/8

P(X=0)+P(X=1)+P(X=2) = 6/8 + 2/8 = 1

This is a probability density function

b. x P(X=x)

0 ----0.2

1 ----0.1

2 ----0.35

3 ----0.17

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.2 + 0.1 + 0.35 + 0.17

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.65 + 0.17 = 0.82 ≠ 1

Therefore this is NOT a probability density function

c. x P(X=x)

0---- 9/10

1 ---- −3/10

2 ---- 3/10

3 ---- 1/10

Since P(X=1) is not between 0 and 1

Therefore this is NOT a probability density function

d. x P(X=x)

0 ----0.06

1 ----0.01

2 ----0.07

3 ----0.86

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.06 + 0.01 + 0.07 + 0.86

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.14 + 0.86 = 1

Therefore this is a probability density function

e. x P(X=x)

0 ----1/2

1 ----1/8

2 ----1/4

3 ----1/8

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/8 + 1/4 + 1/8

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/2 = 1

Therefore this is a probability density function

f. x P(X=x)

0 ----1/10

1 ----1/10

2 ----3/10

3 ----1/5

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/10 + 1/10 + 3/10 + 1/5

P(X=0)+P(X=1)+P(X=2)+P(X=3) = 2/10 + 5/10 = 7/10 ≠ 1

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal([tex]\mu=0.753,\sigma^{2} =0.004^{2}[/tex])

The z-score probability distribution for normal distribution is given by;

                                Z  =  [tex]\frac{X-\mu}{\sigma} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 0.753 inch

           [tex]\sigma[/tex] = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X [tex]\leq[/tex] 0.75 inch)

      P(X < 0.753) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.753-0.753}{0.004} } }[/tex] ) = P(Z < 0) = 0.50

      P(X [tex]\leq[/tex] 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.75) = 1 - P(Z < 0.75)

                                                             = 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the  lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X [tex]\leq[/tex] 0.74 inch)

      P(X < 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z < -0.75) = 1 - P(Z [tex]\leq[/tex] 0.75)

                                                            = 1 - 0.7734 = 0.2266

      P(X [tex]\leq[/tex] 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -3.25) = 1 - P(Z < 3.25)

                                                             = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

      P(X > 0.76) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] > [tex]\frac{0.76-0.753}{0.004} } }[/tex] ) = P(Z > -1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)

                                                            = 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

      P(X < 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z < -3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                                            = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

Using the normal distribution, it is found that there is a

a) 0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

b) 0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

c) 0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

d) 0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

This probability is the p-value of Z when X = 0.753 subtracted by the p-value of Z when X = 0.75.

X = 0.753:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.753 - 0.753}{0.004}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5

X = 0.75:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]

[tex]Z = -0.75[/tex]

[tex]Z = -0.75[/tex] has a p-value of 0.2266

0.5 - 0.2266 = 0.2734

0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

Item b:

This probability is the p-value of Z when X = 0.75 subtracted by the p-value of Z when X = 0.74, hence:

X = 0.75:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]

[tex]Z = -0.75[/tex]

[tex]Z = -0.75[/tex] has a p-value of 0.2266

X = 0.74:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]

[tex]Z = -3.25[/tex]

[tex]Z = -3.25[/tex] has a p-value of 0.0006.

0.2266 - 0.0006 = 0.226

0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 0.76, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.76 - 0.753}{0.004}[/tex]

[tex]Z = 1.75[/tex]

[tex]Z = 1.75[/tex] has a p-value of 0.9599.

1 - 0.9599 = 0.0401

0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

Item d:

This probability is the p-value of Z when X = 0.74, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]

[tex]Z = -3.25[/tex]

[tex]Z = -3.25[/tex] has a p-value of 0.0006.

0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

A similar problem is given at https://brainly.com/question/24663213

Answer:

The base is 6cm

Step-by-step explanation:

Using [tex]\frac{1}{2}[/tex]bh=area, you just plug in 6 for h and 18 for area and solve. The formula should be 3b=18, which equals 6

The solution is Option D.

The length of the base of the triangle is B = 6 cm

A triangle is a plane figure or polygon with three sides and three angles.

A Triangle has three vertices and the sum of the interior angles add up to 180°

Let the Triangle be ΔABC , such that

∠A + ∠B + ∠C = 180°

The area of the triangle = ( 1/2 ) x Length x Base

For a right angle triangle

From the Pythagoras Theorem , The hypotenuse² = base² + height²

if a² + b² = c² , it is a right triangle

if a² + b² < c² , it is an obtuse triangle

if a² + b² > c² , it is an acute triangle

Given data ,

Let the area of the triangle be represented as A

Now , the value of A = 18 cm²

Let the base of the triangle be B

Let the height of the triangle be H = 6 cm

Now , area of the triangle = ( 1/2 ) x Length x Base

Substituting the values in the equation , we get

18 = ( 1/2 ) x B x 6

18 = 3B

Divide by 3 on both sides of the equation , we get

B = 6 cm

Hence , the base length of the triangle is B = 6 cm

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Answer:

Reese read 26 pages on Sunday.

Step-by-step explanation:

If she read twice the amount on Saturday than she read on Sunday, then you can replace the days with variables. 2x for Saturday and x for Sunday. That leaves you with 2x + x = 78. This can be simplified to 3x = 78. We can solve this by dividing both sides by 3. This leaves you with x = 26. If x represents the pages read on Sunday then Reese read 26 pages on Sunday.

Answer:

11.49 dogs/ square mile

Step-by-step explanation:

Population density of any place is number of living being living in that area per unit area.

Population density = number of living being/ area of the place

given

number of dogs= 43,600,000

area = 3,794,083 square miles

population density of domesticated dogs = 43,600,000 /3,794,083

                                                              = 11.4915 dogs/ square mile

population density of domesticated dogs, rounded to the nearest 10th

                                                              is  11.49 dogs/ square mile.

Answer:

A. When its sides are congruent

Step-by-step explanation:

By definition a square is a four sided polygon (shape) which has four right angles and four sides of equal length.

So, when a rectangle has all congruent sides, its a square because its a rectangle so it has four right angles, and congruent sides are sides of the same length.

B is incorrect because a rectangle can have four right angles but have differing side lengths

C is incorrect because again a rectangle can have parallel sides but have differing side lengths

D is incorrect because it can have convex angles but again have differing side lengths

When sides of rectangle are congruent, then it is a square. Therefore, the correct answer is option A.

A rectangle is a closed two-dimensional figure with four sides. The opposite sides of a rectangle are equal and parallel to each other and all the angles of a rectangle are equal to 90°.

A square is a quadrilateral with four equal sides. There are many objects around us that are in the shape of a square. Each square shape is identified by its equal sides and its interior angles that are equal to 90°.

If all the sides of rectangle are congruent, then it is square.

Therefore, the correct answer is option A.

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Answer:

y>5, x+y underlined<12

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

As a simplified fraction, the probability that the first randomly taken candy will be red and second will be red again is [tex]\frac{2}{9}[/tex]

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the candies are selected is not important. So we use the combinations formula to solve this question.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Desired outcomes:

2 candies from a set of 5. So

[tex]D = C_{5,2} = \frac{5!}{2!(5-2)!} = 10[/tex]

Total outcomes:

2 candies from a set of 10. So

[tex]T = C_{10,2} = \frac{10!}{2!(10-2)!} = 45[/tex]

Probability:

[tex]p = \frac{D}{T} = \frac{10}{45} = \frac{2}{9}[/tex]

As a simplified fraction, the probability that the first randomly taken candy will be red and second will be red again is [tex]\frac{2}{9}[/tex]