Jenny’s mom is 36 years old. What percent of Jenny’s mother’s age is Jenny’s great-grandmother’s age (90)?

in the next several games , the pitcher threw a total of 384 pitches and used a fastball 240 times. What decimal should juanita

umka21 [38]

Divide the number of fast balls by the total number of pitches:

240 / 384 = 0.625

7 0

3 years ago

If the square root of the length of the hypotenuse of a right triangle is 2 units, what is the sum of the squares of the length

tatyana61 [14]

the answer is 4 the reason is that the formula is c^2=a^2+b^2,

now if c=2,then 4=a^2+b^2

6 0

3 years ago

12. Higher Order Thinking Find all the

andrew-mc [135]

Answer:

38= 2×19

39= 3×13

40= 2×2×2×5

hcf = highest common factor = 1

if you can't seen any common factor then write 1 , it's common in all.

if all the number divide by same number, then it's common in factor

3 0

3 years ago

One water quality standard for water that is discharged into a particular type of stream or pond is that the average daily water

musickatia [10]

Answer:

$t=\frac{18.15-18}{\frac{3.134}{\sqrt{6}}}=0.117$

$p_v =P(t_{5}>0.117)=0.456$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the mean is not significantly higher than 18 at 10% of signficance.

Step-by-step explanation:

Data given and notation

16.8 21.5 19.1 12.8 18.0 20.7

We can calculate the mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And the results are:

$\bar X=18.15$ represent the sample mean

$s=3.134$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =18$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 18, the system of hypothesis would be:

Null hypothesis: $\mu \leq 18$

Alternative hypothesis: $\mu > 18$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{18.15-18}{\frac{3.134}{\sqrt{6}}}=0.117$

P-value

The degrees of freedom are given by:

$df = n-1 = 6-1= 5$

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

$p_v =P(t_{5}>0.117)=0.456$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the mean is not significantly higher than 18 at 10% of signficance.

3 0

3 years ago

Travis jogged 3 7/12 Miles in the morning. In the afternoon he jogged 2 5/12 Miles . How many fewer miles did Travis jog in the

lana66690 [7]

He traveled 1 1/6 because if you subtract, you get 1 and 2/12, but when you reduce it, it then is 1 1/6.

Hope this helps!

3 0

3 years ago