One state leads the country in tart cherry production, producing 83 out of every 100 tart cherries each year.

Use the Change of Base Formula to evaluate log3 6. Round to the nearest thousandth if necessary.

jeka94

1.556302501 or 1.556

hope this helps

6 0

2 years ago

George buys some first class and second class stamps to put on some letters. Each first class stamp costs 67p, and each second c

Dafna1 [17]

Answer:

£191.36

Step-by-step explanation:

Sum the parts of the ratio, 1 + 4 = 5 parts

Divide 320 by 5 to find the value of one part of the ratio.

320 ÷ 5 = 64 ← value of 1 part of the ratio , thus

4 parts = 4 × 64 = 256

Thus he bought 64 first class and 256 second class.

Cost = (64 × £0.67) + (256 × £0.58)

= £42.88 + 148.48

= £191.36

3 0

2 years ago

A new mixture of self-tanning & moisturizing lotions for everyday use is being developed. This mixture is made by mixing 90

butalik [34]

I don't know but i need points so thanks

7 0

3 years ago

$1080 in the ratio of 1:3:5

MatroZZZ [7]

Divide 1080 by 1+3+5 which is 9
120 is 1
360 is 3
600 is 5
120:360:600

8 0

3 years ago

Read 2 more answers

A business school has a goal that the average number of years of work experience of its MBA applicants is more than three years.

gizmo_the_mogwai [7]

Answer:

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

$p_v =P(z>0.280)=0.390$

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=3.1$ represent the sample mean

$\sigma=\sqrt{6}$ represent the sample standard deviation for the sample

$n=47$ sample size

$\mu_o =3$ represent the value that we want to test

$\alpha$ 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.

We need to conduct a hypothesis in order to check if the mean is higher than 3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3$

Alternative hypothesis: $\mu > 3$

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280$

P-value

Since is a one side test the p value would be:

$p_v =P(z>0.280)=0.390$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.

3 0

3 years ago