Solve for x. Show each step of the solution.

Mr. Wang is buying a $15 box of trail mix at Whole Foods, where tax is 6%. If Mr. Wang has a coupon for 10% off the price of any

algol13

Answer:

$47 plus whole foods market

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the answer to > solve 4cos (theta/3) -2 = 0, over [0, 4pi)

Vesna [10]

Hello from MrBillDoesMath!

Answer:

@ = pi/3 (or 60 degrees) or @ = 7 pi/3 (or 420 degrees)

Discussion:

Let "@' denote the angle "theta". We are asked to find @ in the interval [0, 4 pi)

where

4cos(@) - 2 = 0. Adding 2 to both sides

4 cos(@) - 2 +2 = 2 =>

4 cos(@) = 2 Divide both sides by 4

cos(@) = 2/4 = 0.5

This implies that @ = pi/3 (or 60 degrees) or @ = (pi/3 + 2pi) = 7 pi/3 (or 420 degrees)

Thank you,

MrB

5 0

3 years ago

What is the value of this expresion <br> 4(-6^2+ 3) - 2-1/2

lesya [120]

Answer: -134.5

Step-by-step explanation:

4(-6^2+3) - 2 - 1/2 Exponents first

4(-36 +3) - 2 - 1/2 Now solve the parentheses

4(-33) -2 -1/2

-132- 2 -1/2

-134 - 1/2 = -134.5

7 0

3 years ago

You are hiking in Marin, and stop to stand on the edge of a cliff overlooking the Pacific ocean. You see a sailboat down in the

Bas_tet [7]

Answer:

Step-by-step explanation:

on the edge of a cliff overlooking the Pacific ocean. You see a sailboat down in the water below you. The angle of depression to the sailboat is 28◦ . A quick tells you that the cliff you’re standing on is 150ft. How far away from the boat are you, in feet, rounded to one decimal place?(30pts)

We solve using the Trigonometric function of TANGENT

tan 28 = 150/x

6 0

3 years ago

Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n

Oksanka [162]

Answer:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

$\hat p=\frac{209}{377}=0.554$ estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0

3 years ago