The function in this problem should be: f(x) =x² + 12x + 6

y = x² + 12x + 6
y - 6 = x² + 12x

x² ⇒ x * x
12x ⇒ 2*6*x
missing number is 6² = 36

y - 6 + 36 = x² + 12x + 36

(x+6)(x+6) ⇒ x(x+6)+6(x+6) ⇒ x² + 6x + 6x + 36 = x² + 12x + 36

y + 30 = x² + 12x + 36
y = (x+6)² - 30

Choice is D. 36,-36

5 0

3 years ago

Emily is entering a bicycle race for charity. Her mother pledges $0.70 for every 0.5 mile she bikes. If Emily bikes 10 miles, ho

Jlenok [28]

Answer:

Emily's mother will donate $20

Step-by-step explanation:

0.5x20=10

7 0

3 years ago

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Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg

Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let X = number of college women on a diet.

The probability of a woman being on diet is, P (X) = p = 0.70.

The sample of women selected is, n = 267.

The random variable thus follows a Binomial distribution with parameters n = 267 and p = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions ( $\hat p$ ) follows a Normal distribution.

The mean of this distribution is:

$\mu_{\hat p} = p = 0.70$

The standard deviation of this distribution is: $\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028$

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

$P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z$

**Use the z-table for the probability.

$P(\hat p \geq 0.75)=1-P(Z$

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0

3 years ago