Find the slope of the line that passes through the pair of points.<br> (4,8), (8, 11)

PLEASE HELP Find the length of PQ in circle R to the nearest hundredths

Vladimir [108]

There's no way to do that without seeing the drawing or hearing a lot more information. For one thing, we don't even know the size of the circle yet.

5 0

2 years ago

Researchers fed mice a specific amount of Dieldrin, a poisonous pesticide, and studied their nervous systems to find out why Die

Elodia [21]

Answer:

Step-by-step explanation:

Part A

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

6 0

3 years ago

Lily bought 20.24 pounds of grapefruit. The lightest grapefruit weighed 1.2 pounds. The heaviest grapefruit weighed 1.8 pounds.

dexar [7]

Given:

Lily bought 20.24 pounds of grapefruit.
The lightest grapefruit weighed 1.2 pounds.
The heaviest grapefruit weighed 1.8 pounds.

To Find:

An estimate of the number of grapefruits she bought.

Answer:

The best estimate is that Lily bought 13 grapefruits.

Step-by-step explanation:

Given that the lightest grapefruit weighed 1.2 pounds and the heaviest one weighed 1.8 pounds, we can take the mean or average of these two weights to estimate the weight of any of the grapefruits that she bought.

That is, we can say that each grapefruit must weigh something close to the average value.

The average can be found as

$\frac{1.2+1.8}{2}=1.5$