Find the coordinates of the vertex of the following parabola algebraically. Write your

The average expenditure on Valentine's Day was expected to be$100.89 (USA Today, February 13, 2006). Do male and femaleconsumers

stealth61 [152]

Answer:

(a) $62.16

(b) Male: $15.00

Female: $10.06

(c) Confidence Interval for male expenditure is ($106.40, $136.40)

Confidence interval for female expenditure is ($49.18, $69.30)

Step-by-step explanation:

(a) Male expenditure

Sample mean = $135.67, sd=$35, n=40, Z=2.576

Population mean = sample mean - (Z×sd)/√n = 135.67 - (2.576×35)/√40 = 135.67 - 14.27 = $121.40

Female expenditure

Sample mean= $68.64, sd=$20, n=30, Z=2.576

Population mean = 68.64 - (2.576×20)/√30 = 68.64 - 9.40 = $59.24

$121.40 - $59.24 = $62.16

(b) Male: Error margin = (t-value × sd)/√n

Degree of freedom = n-1 = 40-1= 39. t-value corresponding to 39 degrees of freedom and 99% confidence level is 2.708

Error margin = (2.708×35)/√40 = 94.78/6.32 = $15.00

Female

Degrees of freedom = n-1 = 30-1 = 29. t-value is 2.756

Error margin = (2.756×20)/√30 = 55.12/5.48 = $10.06

(c) Male

Confidence Interval (CI) = (mean + or - error margin)

CI = 121.4 + 15.00 = $136.40

CI = 121.4 - 15.00 = $106.40

Confidence Interval is ($106.40, $136.40)

Female

CI = 59.24 + 10.06 = $69.30

CI = 59.24 - 10.06 = $49.18

Confidence Interval is ($49.18, $69.30)

7 0

3 years ago

I need help! someone please answer asap!

kirza4 [7]

For your question the answer is C.

Hope this helps:)

5 0

3 years ago

Read 2 more answers

Can someone help me please , this geometry

sashaice [31]

Answer:

x = HJ/5 + 3/5

HJ = 5x - 3 = 5 (11) - 3 = 55 - 3 = 52

JK = 8x - 9 = 8(11) - 9 = 88 - 9 = 79

5 0

1 year ago

certain computer virus can damage any file with probability 35%, independently of other files. Suppose this virus enters a folde

Ksivusya [100]

Answer:

0.623 is the probability that between 800 and 850 files get damaged.

Step-by-step explanation:

We are given the following information:

We treat virus can damage computer as a success.

P( virus can damage computer) = 35% = 0.35

The conditions for normal distribution are satisfied.

By normal approximation:

$\mu = np = 2400(0.35) = 840\\\sigma = \sqrt{np(1-p)} = \sqrt{2400(0.35)(1-0.35)} = $$23.36$

We have to evaluate probability that between 800 and 850 files get damaged.

$P(800 \leq x \leq 850) = P(\displaystyle\frac{800 - 840}{23.36} \leq z \leq \displaystyle\frac{850-840}{23.36}) = P(-1.712 \leq z \leq 0)\\\\= P(z \leq 0.428) - P(z < -1.712)\\= 0.666 - 0.043 = 0.623 = 62.3\%$

$P(800 \leq x \leq 850) = 62.3\%$

0.623 is the probability that between 800 and 850 files get damaged.

8 0

3 years ago

The type of probability that you think will happen

PolarNik [594]

If I know what you're asking, the answer is theoretical probability.

8 0

3 years ago