Use the discriminant to describe the roots of each equation. Then select the best description.

1 answer:

4 0

Let's find the discriminant of x^2+9x+14=0. Here, a=1, b=9 and c=14.

The discriminant is b^2-4ac. Substituting the above numeric values,
9^2-4(1)(14) = 81-56 = 25

The sqrt of 25 is 5. Thus, your polynomial has two unequal, real roots.

Off the point example: If the discriminant were zero, your poly would have two real, equal roots.

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A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

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Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s as follows:

$z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57$

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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Yvonne is interested in testing the effectiveness of a new flea medication on dogs, so she divided some dogs with fleas into two

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Actually, the real answer is "NO, BECAUSE GROUP B SHOULD BE THE CONTROL GROUP". Since Yvonne gave the flea medication to the group A, and we need a control group. If you don't know what it's the control group, it's the group that they don't put the experiment on, just to see if the group(s) being experimented, had any positive or negative changes. In a short summary they need group B to stay like they are to compare it to the experimented group (group A)

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