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hoa [83]
3 years ago
10

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

Mathematics
2 answers:
wel3 years ago
8 0

Answer:81%

Step-by-step explanation:

MaRussiya [10]3 years ago
3 0

Answer:

60% tiger trout

Step-by-step explanation:

First, do 22+18=  40

Then, when doing % its always out of 100. So, 100 - 40= 60%

Which means, there were 60% Tiger trout.                                                                                                                                                            

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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
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