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Please help me on this one. I don't know the equation to solve it. I really need help!!

BigorU [14]

I can't see the question it's blurry

4 0

3 years ago

The Bishop's Peak 4-H club is having its annual fundraising dinner. Adults pay $35 apiece and children pay $25 apiece. If the nu

vodka [1.7K]

35a + 25c = 2660
a = 2c

35(2c) + 25c = 2660
70c + 25c = 2660
95c = 2660
c = 2660/95
c = 28 <=== there were 28 children tickets

a = 2c
a = 2(28)
a = 56 <== there were 56 adult tickets

4 0

3 years ago

If x=square root of 6 and y^2=12 then 4/xy=

vova2212 [387]

Y² = 12
y = 2√3

x = √6

Now, 4 /xy
= 4/2√3√6
= 2 / 3√2
= √2/3

In short, Your Answer would be √2 / 3

Hope this helps!

5 0

4 years ago

Read 2 more answers

A gardener has 27 pansies and 36 daisies. He plants an equal number of flowers in each row. What is the greatest possible number

Serggg [28]

There would be a a row of seven and would be 9 in each row

7 0

4 years ago

Read 2 more answers

A random sample of 16 students selected from the student body of a large university had an average age of 25 years. We want to d

kenny6666 [7]

Answer:

$z=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

$p_v =2*P(Z>2)=0.0455$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean differs from 24 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=25$ represent the sample mean

$\sigma=2$ represent the sample population deviation for the sample

$n=16$ sample size

$\mu_o =24$ represent the value that we want to test

$\alpha=0.05$ 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 true mean is different from 24, the system of hypothesis would be:

Null hypothesis: $\mu = 24$

Alternative hypothesis: $\mu \neq 24$

If we analyze the size for the sample is < 30 but 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:

$z=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(Z>2)=0.0455$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean differs from 24 at 5% of significance

6 0

3 years ago