9/6=x/10 what is x? How to figure it out.

How many of the numbers from the set $\{1,\ 2,\ 3,\ldots,\ 50\}$ have a perfect square factor other than one?

Novay_Z [31]

Answer:

2^2 ( 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44,48

3^2(1,2,3, 5) = 9, 18, 27, 45

5^2(1, 2) = 25, 50

7^2 = 49

19 numbers

Step-by-step explanation:

5 0

3 years ago

15w2+62w+63=0 <br><br><br> Can you help me

anzhelika [568]

Answer:

w=-7/3 and w=-9/5

Step-by-step explanation:

Given: 15w^2+62w+63=0

Factor: (3w+7)(5w+9)=0

Zero Product Property: 3w+7=0 and 5w+9=0

Solve each equation: w=-7/3 and w=-9/5

Let me know if you would like me to explain more.

7 0

2 years ago

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

What is the answer <br> How do you solve it

stich3 [128]

Answer:

-50y+(-80)

Step-by-step explanation:

3 0

2 years ago

Hospital records show that a certain surgical procedure takes on an average 111.6 minutes with a standard deviation of 2.8 minut

Arisa [49]

Answer: