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3 years ago

Need to know the answer

Mathematics

2 answers:

Rzqust [24]3 years ago

8 0

Here you go I hope this helps.

zmey [24]3 years ago

4 0

Answer:

x = 3

Step-by-step explanation:

use distribution to multiply everything in the parenthesis by -5.

(-5x - 40) = -55

add 40 to both sides to isolate the variable which in this case is x

-5x = -15

simplify

x = 3

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For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

5 0

3 years ago

Helpppppppp pleaseeee!!!!!!!!

SashulF [63]

Answer:

29.5 square inches

Step-by-step explanation:

Let's divide the shape into 3 parts to make it easy for us to find the area.

- a wide triangle

- a triangle

- a shorter rectangle

For the wider rectangle, the area is;

A1 = 8 × 2 = 16 in²

For the triangle, the area is;

A2 = ½(7 - 2) × 3

A2 = 7.5 in²

For the shorter rectangle, the area is;

A3 = 3 × 2

A3 = 6 in²

Total area = A1 + A2 + A3 = 16 + 7.5 + 6 = 29.5 square inches

8 0

3 years ago

Let e f (x) = x2 – 2x - 4 What is the average rate of change for the quadratic function from X = -1 tox to x = 42 Enter your ans

sammy [17]

Answer:

The average rate of change of the given function

A(x) = 1

Step-by-step explanation:

Step(i):-

Given function f(x) = x² - 2x -4

And given that x = a = -1 and x=b = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

Step(ii):-

f(x) = x² - 2x -4

f(-1) = (-1)² - 2(-1) -4 = 1+2-4 = -1

f(4) = 4² -2(4) -4 = 16 -12 = 4

The average rate of change of the given function

$A(x) = \frac{f(b)-f(a)}{b-a}$

$A(x) = \frac{4-(-1)}{4-(-1)} = \frac{5}{5} = 1$

final answer:-

The average rate of change of the given function

A(x) = 1

5 0

3 years ago

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4 years ago

In 2-3 sentences describe how you use substitution to solve a system of linear equations?

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3 years ago