(PLEASE ANSWER + BRAINLIEST ANSWER!!)

Find each Unit rate:

raketka [301]

The unit rate is 2.5 meters in 1 second

40/16=2.5

5 0

2 years ago

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Can someone help me with this

sammy [17]

Answer:

Step-by-step explanation:

4 0

2 years ago

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Find each difference. If you get stuck, consider drawing a number line diagram. HELP ASAP

liq [111]

1.) 5
2.) 13
3.) -350
4.) 150
5.) -520
6.) 200
7.) 5
8.) -9
9.) -3
10.) 13
hope this helped

4 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

Pls help!!!!!!!<br> 20-27

xeze [42]

Answer:

The probabilities would be 4/11 and 7/11 respectively.

Step-by-step explanation:

1) Total length of the given word i.e. Mathematics = 11

2) Number of vowels in the given word = 4 i.e. a,e,a,i

3) Number of consonants in the given word = 7 i.e. m,t,h,m,t,c,s

4) The probability of choosing a vowel would be 4/11 = 0.3636..

5) The probability of choosing a consonant would be 7/11 = 0.6363...

Step-by-step explanation:

8 0

1 year ago