MATHEMATICS HIGH SCHOOL

Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to increase daily sales. To determine the effectiveness of the advertising campaign, a sample of 49 days of sales were taken. They found that the average daily sales were $6,300 per day. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice A. Reject the null hypothesis and conclude that the mean is equal to $6,000 per day.
B. Fail to reject the null hypothesis.
C. Reject the null hypothesis and conclude the mean is lower than $6,000 per day.
D. Reject the null hypothesis and conclude the mean is higher than $6,000 per day.

Answers

Answer 1

Answer:

Answer:

Option B) Fail to reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $6,000

Sample mean, $\bar{x}$ = $6,300

Sample size, n = 49

Alpha, α = 0.01

Population standard deviation, σ = $1,000

First, we design the null and the alternate hypothesis

$H_(0): \mu = 6000\text{ dollars per week}\nH_A: \mu > 6000\text{ dollars per week}$

We use one-tailed(right) z test to perform this hypothesis.

Formula:

$z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }$

Putting all the values, we have

$z_(stat) = \displaystyle(6300 - 6000)/((1000)/(√(49)) ) = 2.1$

Now, $z_(critical) \text{ at 0.01 level of significance } = 2.33$

Since,

$z_(stat) < z_(critical)$

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that sales have not increased as a result of the advertising campaign

Option B) Fail to reject the null hypothesis.

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Dad is 4 times as old as his son Jim. In 10 years, Dad's age will be 20 years more than twice Jim's age. How old is Jim?If D = dad's age now and J = Jim's age now, which of the following systems could be used to solve the problem? a. D = 4J and D + 10 = 2J + 20 b. D = 4J and D + 10 = 2(4J + 10) + 20 c. D = 4J and D + 10 = 2(J + 10) + 20
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Yesterday, the snow was 2 feet deep in front of Archie’s house. Today, the snow depth dropped to 1.6 feet because the day is so warm. What is the percent change in the depth of the snow?

Answers

earlier snow depth = 2 feet
later snow depth = 1.6 feet

change
2 - 1.6
0.4 feet

percentage change
(0.4 / 2) * 100
20 %

The correct answer of this question is 20%.
The given values are the initial depth of snow which is 2 feet and the final depth which is 1.6 feet.
The solution of the problem is:
2 ft. - 1.6 ft. = 0.4 ft.
0.4 ft. / 2 ft. = 0.2
To calculate for the percentage,
0.2 x 100% = 20%

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. x^2 + 3 = -4x

A. 1,3
B. -1,-3
C. 1,-3
D. -1, 3

Answers

Hello,

x²+4x+3=0
x=(-4-2)/2=-3 or x=(-4+2)/2=-1

Answer B (-1,-3)

Three workers sort 66 boxes in a day. One of the workers is three times more productive than the second worker and twice as productive as the third worker. How many boxes did the second most productive worker sort?

Answers

1 worker $=x$

2 worker $= (worker1)/(2)= (x)/(3)$

3 worker $= (worker1)/(3)= (x)/(3)$

so:

$66=x+ (x)/(3)+ (x)/(2)$

$66=x+(2x)/(6)+ (3x)/(6)$

$66*6=6x+2x+3x$

$396=11x$

$x=36$

so:

worker $1$ sorted $36$ boxes
worker $2$ sorted $18$ boxes
worker $3$ sorted $12$ boxes

Kayden is a stunt driver. One time, during a gig where she escaped from a building about to explode(!), she drove at a constant speed to get to the safe zone that was 160 meters away. After 3 seconds of driving, she was 85 meters away from the safe zone. Let D represent the distance (in meters) from the safe zone after t seconds. Complete the equation for the relationship between the distance and number of seconds.

Answers

$d=160-25t$

Step-by-step explanation:

Let $d$ be the distance between Kayden and safe zone at any time.

It is given that initially $d=160$

Lrt her speed be $v$

Since the speed is constant,the distance covered by her in $t$ seconds= $v* t$

So,the distance between her and safe zone after $t$ seconds is $160-vt$

It is given that after $3$ seconds, $d=85$

So, $160-3v=85$

$3v=75$

$v=25$

So, $d=160-3t$

Answer:

y=-25x+160

Step-by-step explanation:

Write -2.55 as a mixed number

Answers

Answer: 2.55 as a mixed number is 51/20

The airline industry defines an on-time flight as one that arrives within 15 minutes of its scheduled time. The following table shows the number of on-time and late flights leaving New York and arriving in Miami between November 1 and December 31, 2018, by airlines: Airlines On-Time Late Total United 254 72 326 Delta 292 65 357 American 235 58 683 Total 781 195 a. What is the probability that a randomly selected flight was Delta and was late? b. What is the probability that a randomly selected flight was United or was on-time? c. Given the flight was late, what is the probability that it was from American? d. Given the flight was from Delta, what is the probability that it was late? e. Construct a probability tree for these probabilities.

Answers

a) The probability is $\frac{65}{357}$.

b) The probability is $\frac{781}{1000}$.

c) The probability is $\frac{58}{195}$.

d) The probability is $\frac{65}{357}$.

e) We can construct a probability tree for these probabilities. Below is the probability tree:probability tree for airlines. Therefore, the above figure is the probability tree for airlines.

The probability that a randomly selected flight was Delta and was late is $\frac{65}{357}$.Therefore, the probability is $\frac{65}{357}$.

The probability that a randomly selected flight was United or was on-time is $\frac{781}{1000}$.Therefore, the probability is $\frac{781}{1000}$.

Given the flight was late, the probability that it was from American is $\frac{58}{195}$.Therefore, the probability is $\frac{58}{195}$.

Given the flight was from Delta, the probability that it was late is $\frac{65}{357}$.Therefore, the probability is $\frac{65}{357}$.

We can construct a probability tree for these probabilities. Below is the probability tree:probability tree for airlines. Therefore, the above figure is the probability tree for airlines.

Learn more about Probability

brainly.com/question/30034780

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