Kendrick caught 22 rainbow trout and 18 tiger trout while fishing last month. What percent of the trout were tiger trout?

If she was vaping every 15 minutes, assuming she was awake from 7 am- 11 pm (16 hours) how many times in a day would she vape?

Ray Of Light [21]

Answer:

64 times

Step-by-step explanation:

16 hours to minutes: 16*60 = 960 minutes

vaping every 15 minutes: 960/15 = 64

8 0

2 years ago

I need help on the yes and no one

lutik1710 [3]

Hello from MrBillDoesMath!

Answer:

10a. YES

10b. YES

10c. NO

10d . YES

Discussion:

If x = 12

10a. Does (3/4) 12 = 9? Does (36/4) = 9 = 9? YES

10b. Does 3x = 36? Does 3(12) = 36 = 36? YES

10c. Does 5x = 70? Does 5(12) = 60 = 70? NO

10d. Does x/3 = 4? Does 12/3 = 4 = 4 ? YES

Thank you,

MrB

5 0

3 years ago

Records at an insurance company showed that the first 30 customers made an average of 5 claims each through their lifetime. Afte

umka2103 [35]

The average of the 75 customers will be 6.4.

<h3>How to calculate the value?</h3>

The first 30 customers made an average of 5 claims each. The total claims will be:

= (5 × 30) = 150

When 75 more people were assessed, the average was 6. Therefore, the total claims will be:

= (75 + 30) × 6

= 105 × 6

= 630

Therefore, the average of the 75 customers will be:

= (630 - 150)/75

= 480/75

= 6.4

Learn more about mean on:

brainly.com/question/1136789

#SPJ1

8 0

2 years ago

1 Which relation is not a function?

Karo-lina-s [1.5K]

3 because the x value has been repeated

6 0

3 years ago

Read 2 more answers

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of

Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

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

t 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 to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

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

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

6 0

3 years ago