Help me i need this on sep 16 2019

Travis recieved a box of chocolate. He wants to eat one row of chocolate a day.there are 4 chocolates in a row .how many chocola

Korolek [52]

Answer:

24

Step-by-step explanation:

if he eats 4 each day for 6 days, he will eat 24

4 × 6 = 24

3 0

3 years ago

A 12 foot ladder leans against a wall, making an angle of 40° with the horizon. How far is the base of the ladder from the wall?

olga_2 [115]

the base of the ladder is 18ft away from the wall

6 0

3 years ago

Graph the function f(x) = 1/2(2)^x

xxMikexx [17]

Answer; Down

Step-by-step explanation:The graph of the squaring function has the shape of a parabola that opens up. Its vertex is at (0,0). The graph of F(x) = (1/2)x^2 has the same shape, but is compressed vertically by a factor of (1/2).

3 0

2 years ago

How many square roots dose the number 20 have

ss7ja [257]

Answer:

The awnser is 4.472135955

7 0

3 years ago

Suppose a poll is taken that shows that 765 out of 1500 randomly selected, independent people believe the rich should pay more

Zanzabum

Answer:

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

Step-by-step explanation:

Data given and notation

n=1500 represent the random sample taken

X=765 represent the successes

$\hat p=\frac{765}{1500}=0.51$ estimated proportion of successes

$p_o=0.5$ 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)

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.51 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1500}}}=0.775$

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:

$p_v =P(z>0.775)=0.219$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.5

8 0

3 years ago