The high temperature is 35 degrees and the low is -2 degrees. What is the difference between the high temp and low temp?

Simplify the variable expression by evaluating its numerical parts and enter your answer m - 22 + 37 - 8

skad [1K]

M-22+29

I think I hope this helps

5 0

3 years ago

The experimental probability of a coin landing on heads is 7/12. If the coin landed on tails 30 times, find the number of tosses

nordsb [41]

Answer: 72 tosses

Explanation: If the probability of landing heads is 7/12, the probability of landing tails is 5/12. So then you do cross multiplication, where you multiply the 12 and the 30, equalling 360. Then divide that by 5, equalling 72.

4 0

4 years ago

I had 6 bananas and 4 apples. A hungry baboon ate two of the fruits before being chased by a dog.find probability that the baboo

Wittaler [7]

Answer:

two bananas because he's a monkey

Step-by-step explanation:

5 0

2 years ago

Maxine wants a necklace 42 beads long of ratio 2:5.

BlackZzzverrR [31]

Answer:

21

Step-by-step explanation:

8 0

3 years ago

A husband and wife ted and suzanne share a digital music player that has a feature that randomly selects which song to play. A t

seraphim [82]

Answer:

$z=\frac{0.733 -0.5}{\sqrt{\frac{0.5(1-0.5)}{30}}}=2.552$

$p_v =2*P(z>2.552)=0.0107$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 songs selectd at random differs from 0.5 .

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.

Step-by-step explanation:

Data given and notation

n=30 represent the random sample taken

X=22 represent the number of songs selectd at random

$\hat p=\frac{22}{30}=0.733$ estimated proportion of songs selectd at random

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

$\alpha$ represent the significance level

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 of random songs selected is 0.5 (random) or no.:

Null hypothesis: $p=0.5$

Alternative hypothesis: $p \neq 0.5$

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:

$z=\frac{0.733 -0.5}{\sqrt{\frac{0.5(1-0.5)}{30}}}=2.552$

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 assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>2.552)=0.0107$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ 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 songs selectd at random differs from 0.5 .

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.

7 0

3 years ago