In the data set below what is the mean absolute deviation 41,56,38,45,55,51,52

Quinn returned home one summer's day to find it sweat-inducingly hot! He turned the air conditioner on and fell asleep. The room

Alenkasestr [34]

Answer:

$f(x)=-0.5x+40$

Step-by-step explanation:

To solve this problem, we need to find the linear function. We know that the constant rate of change is -0.5° Celsius per minute. Also, after 60 minutes the temperature was 10° Celsius. So, we have a one point and the slope of the linear function, let's use the point-slope formula

$y-y_{1} =m(x-x_{1} )\\y-10=-0.5(x-60)\\y=-0.5x+30+10\\y=-0.5x+40$

Where the y-intercept is at (0, 40).

Now, we have two points to graph the relation between minutes and Celsius degrees.

Therefore, the room's temperature as a function of time is

$f(x)=-0.5x+40$

Its graph is attached.

5 0

2 years ago

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Volgvan

Answer: We reject the null hypothesis, and we use Normal distribution for the test.

Step-by-step explanation:

Since we have given that

We claim that

Null hypothesis : $H_0:\mu\geq 50000$

Alternate hypothesis : $H_1:\mu$

There is 5% level of significance.

$\bar{X}=46800\\\\\sigma=9800\\\\n=29$

So, the test statistic would be

$z=\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{46800-50000}{\dfrac{9800}{\sqrt{29}}}\\\\z=-1.75$

Since alternate hypothesis is left tailed test.

So, p-value = P(z≤-2.31)=0.0401

And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.

So, we reject the null hypothesis, and we use Normal distribution for the test.

4 0

2 years ago

Suppose the time required for an auto shop to do a tune-up is normally distributed, with a mean of 102 minutes and a standard de

hammer [34]

Answer:

Step-by-step explanation:

Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean time

s = standard deviation

From the information given,

u = 102 minutes

s = 18 minutes

1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as

P(x > 120 minutes) = 1 - P(x ≤ 120)

For x = 120

z = (120 - 102)/18 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x > 120) = 1 - 0.8413 = 0.1587

2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as

P(x < 66 minutes)

For x = 66

z = (66 - 102)/18 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

P(x < 66 minutes) = 0.02275

4 0

2 years ago

A local university wanted to understand what students prefer to eat during finals. They asked 1,000 students, "Do you prefer chi

kompoz [17]

Given that the person is female, the universe is reduced to 219 + 192 + 119 = 530 people.

The number of women that prefer pizza is 119.

Then the probability is 119/530 *100 = 22.5%

3 0

2 years ago

Read 2 more answers

Sally has just finished her thirty-fifth year with her company and is getting ready to retire. During her thirty-five years, Sal

olya-2409 [2.1K]

The <em>correct answer</em> is:

$19,153.26.

Explanation:

She will receive 42% of her average annual salary. This salary was $45,603.

To find 42% of a number, we first convert the percent to a decimal: 42% = 42/100 = 0.42.

Next we multiply this by the number we're taking the percent of:

0.42(45603) = 19153.26.