(brainliest) please help me out with this graph! :)

WILL MARK BRAINLIEST IF YOU ANSWER NOW!!!

Mila [183]

Answer: C. 3n + 3 = 198

3 times Monday plus 3

3 times the number they sold Monday represents 3n.

3 more represents +3.

198 is just the total.

Hope this helps :)

6 0

4 years ago

The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample me

spin [16.1K]

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ $\sigma^{2}$ ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

$\frac{\alpha}{2}$ = 0.025

t-score = $t_{0.025,11}$ = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± $t.\frac{s}{\sqrt{n} }$

Substituing values:

21.9 ± 2.200. $\frac{7.7}{\sqrt{12} }$

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

$s^{2}$ = $7.7^{2}$

$s^{2}$ = 59.29

α = 0.05

$\frac{\alpha}{2}$ = 0.025

$1-\frac{\alpha}{2}$ = 0.975

$\chi^{2}_{0.025,11}$ = 21.92

$\chi^{2}_{0.975,11}$ = 3.816

Note: To find the values for $\chi^{2}_{\alpha/2,n-1}$ and $\chi^{2}_{1-\alpha/2,n-1}$ , look for them at the chi-square table

The formula to calculate interval:

( $\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}}$ )

are the lower and upper limits, respectively.

Substituing values:

( $\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816}$ )

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ $\sigma^{2}$ ≤ 170.9093

6 0

3 years ago

( URGENT) What are the roots of the polynomial equation x Superscript 4 Baseline + x cubed = 4 x squared + 4 x? Use a graphing c

Finger [1]

Answer:

A.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A particular variety of watermelon weighs on average 22.4 pounds with a standard deviation of 1.36 pounds. Consider the sample m

daser333 [38]

Answer:

a) 22.4 pounds.

b) 0.17 pounds.

c) 0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.

d) c = 22.62

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .