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3 years ago

Answer these two questions.

Mathematics

1 answer:

Jlenok [28]3 years ago

4 0

Hey! Number one is x is less then 60 and number two is 2 is less then x, which is less then 8. Hope this helps:)

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If m∠ABC = 72°, what is m∠DAB? A) 72° B) 108° C) 144° D) 216°

11111nata11111 [884]

The answer would be B
What you would do is 180-72= 108 degrees

3 0

2 years ago

Read 2 more answers

Seven less than a number, divided by 5, is -2

ycow [4]

Answer:

$\frac{x - 7}{5} = 2 \\cross \: multiply \\ x - 7 = 10 \\ x = 17$

hope this helps you

4 0

3 years ago

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

2 years ago

In problems below, (a) identify the independent variable and the dependent variable of each equation (use 't' for the independen

7nadin3 [17]

Step-by-step explanation:

The unanimous Declaration of the thirteen united States of America, When in the Course of human events, it becomes necessary for one people to dissolve the political bands which have connected them with another, and to assume among the powers of the earth, the separate and equal station to which the Laws of Nature

8 0

3 years ago

Find the degree of the algebraic expression: 3x^3 + 3x^2y^2 - y^2

Ghella [55]

Answer:

Degree 4

Step-by-step explanation:

Expression:

3x^3 + 3x^2y^2 - y^2

3x^3 - degree 3
3x^2y^2 - degree 2+2 = 4
y^2 - degree 2

So the expression has degree 4 as per the highest degree of the terms

4 0

3 years ago