Which expression is equivalent to (2g^5)^3/(4h^2)^3

x = 3

-2 | 2.2(3) - 3.3 | = -6.6

-2 | 6.6 - 3.3 | = -6.6

-2 | 3.3 | = -6.6

-2 * 3.3 = -6.6

-6.6 = -6.6

Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

∵ this is a two-tailed test.

Test statistic:

, where p= denotes the population proportion

= signifies the sample proportion

n= represents the sample size

Setting n= 6 and and p=0.08

P-value for the two-tailed test:[2P(Z>|z|)

=2P(Z>|-0.415|)

=2(1-0.6609) [from the z-table]

Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.

This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

Complete question;

A booster club financed pizzas for an end-of-season celebration. The cost was $13 for each pepperoni pizza and $11 for each plain pizza. They purchased a

total of 18 pizzas for the event, totaling $212.

Organize the numbers into an equation that represents this scenario with x being the number of pepperoni pizzas and y being the number of plain pizzas purchased.

Answer:

13x + 11y = 212

Step-by-step explanation:

Two varieties of pizza were purchased: pepperoni and plain. The pepperoni variety cost $13 each while the plain one cost $11 each. In total, they bought 18 pizzas.

The overall expenditure on the pizzas was $212.

total number of pizzas = 18

total amount spent on pizzas = $212

x represents the number of pepperoni pizzas

y represents the number of plain pizzas

The cost of the pepperoni pizzas is derived by multiplying the number of pizzas by their individual price. The same calculation applies for the plain pizzas.

Therefore, by summing the costs of both pizza types, we arrive at the total price for the pizzas.

13x + 11y = 212