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liubo4ka [24]
2 years ago
10

Find the length of AB

Mathematics
1 answer:
Roman55 [17]2 years ago
6 0
If i right understand task.
6/4=1.5
15/x=1.5 => x=15/1.5=10
ab=x+4=14
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A large pizza has diameter 36 cm and a medium pizza has diameter 30 cm. What is the difference in area of the pizzas? Round your
vladimir1956 [14]

Answer:

311 cm^{2}

Step-by-step explanation:

Taking a pizza to be a circle

Area of big pizza be x abd are of small pizza be y

hence difference = x-y

Area of a circle = \pi r^{2} where r is the radius

taking \pi to be 3.14

hence area x = 3.14 (\frac{36}{2} *\frac{36}{2}) = 1017.36

Area y = 3.14 (\frac{30}{2} * \frac{30}{2} )= 706.5

x-y = 1017.36- 706.5 =310.86 cm^{2}

310.86 cm^{2}

6 0
2 years ago
The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an
Genrish500 [490]

Answer:

 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,  

Standard deviation(\sigma) = 7.1

Mean \overline{(y)} = 110

\alpha = 1 - confidence interval = 1 - .98 = .02

z_{\frac{\alpha}{2}} = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = \overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}

                     = 110\pm z_{.01}\frac{7.1}\sqrt{50}

                      = 110\pm 2.33\frac{7.1}\sqrt{50}

       First we take  + sign

   110 +  2.33\frac{7.1}\sqrt{50} = 112.34

now  we take  - sign

110 -  2.33\frac{7.1}\sqrt{50} = 107.66

 We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .

               

5 0
3 years ago
Use the ordinary division of polynomials to find the quotient and remainder when the first polynomial is divided by the second.
Vanyuwa [196]

The quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively

<h3>How to determine the quotient and remainder when the first polynomial is divided by the second?</h3>

The polynomials are given as:

-4w^3 + 5w^2 - 8, w - 3

Set the divisor to 0.

So, we have

w - 3 = 0

Add 3 to both sides

w = 3

Substitute w = 3 in -4w^3 + 5w^2 - 8 to determine the remainder

-4(3)^3 + 5(3)^2 - 8

Evaluate the expression

-71

This means that the remainder when -4w^3 + 5w^2 - 8 is divided by w - 3 is -71

The quotient (Q) is calculated as follows:

Q = [-4w^3 + 5w^2 - 8]/[w - 3]

The numerator can be expressed as follows:

Numerator = -4w^3 + 5w^2 - 8

Subtract the remainder.

So, we have:

Numerator = -4w^3 + 5w^2 - 8 + 71

This gives

Numerator = -4w^3 + 5w^2 + 61

So, the quotient becomes

Q = [-4w^3 + 5w^2 + 61]/[w - 3]

Expand

Q = [(w - 3)(-4w^2 - 7w - 21)]/[w - 3]

Evaluate

Q = -4w^2 - 7w - 21

Hence, the quotient and remainder when the first polynomial is divided by the second are -4w^2 - 7w - 21 and -71 respectively

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

6 0
1 year ago
Macy wants to know if the number of words on a page in her grammar book is generally more than the number of words on a page in
dybincka [34]

yes, because there is a lot of variability in the grammar book data.

6 0
2 years ago
Multiple regression is sometimes used in litigation. In the case of Cargill, Inc. v. Hardin, the prosecution charged that the ca
bogdanovich [222]

Answer:

($2.123 ; $2.149)

Step-by-step explanation:

The prediction interval is expressed as :

Predicted value ± standard Error

Predicted value = $2.136

Standard Error = $0.013

Prediction interval :

Lower boundary = $2.136 - $0.013 = $2.123

Upper boundary = $2.136 + $0.013 = $2.149

($2.123 ; $2.149)

B.) The prediction interval provides a range for which the predicted value or price should fall Given a certain degree of probability. If the true value falls within this interval, then, our prediction would be deemed to have occurred not by chance.

Since the actual price within the predicted price interval, then I agree with the judge's Decison that the price was not artificially depressed.

4 0
2 years ago
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