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timurjin [86]
4 years ago
6

Plzzzzzx anyone help me

Mathematics
1 answer:
Yuki888 [10]4 years ago
8 0

First look at the points and find their x and y values. So we have:

(-3,3), (0,0), (3,3)

Next separate the x values and the y values

x values: -3, 0, 3

y values: 3, 0, 3

The x values are your domain and the y values are are range

So the answer is b

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Answer quick before I fail :)
insens350 [35]

I’m not 100% sure but I think it’s 14/49

5 0
3 years ago
Read 2 more answers
What is the answer for 7-2(8/2)+9
photoshop1234 [79]
7-2( \frac{8}{2} )+9 =

Remove parentheses:

7-2* \frac{8}{2} +9 =

7 - 2 * 4 + 9 =

7 - 8 + 9 =

= -1 + 9

= 8

hope this helps!



6 0
3 years ago
The​ quality-control manager at a compact fluorescent light bulb​ (CFL) factory needs to determine whether the mean life of a la
lara31 [8.8K]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 7463

For the alternative hypothesis,

µ ≠ 7463

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 7463 hours

x = 7163 hours

σ = 1080 hours

n = 81

b) z = (7163 - 7463)/(1080/√81) = - 2.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.02

Recall, population mean is 7463

The difference between sample sample mean and population mean is 7463 - 7163 = 300

Since the curve is symmetrical and it is a two tailed test, the x value for the left tail is 7463 - 300 = 7163

the x value for the right tail is 7463 + 300 = 7763

These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area below z as 0.02

We would double this area to include the area in the right tail of z = 2.5. Thus

p = 0.02 × 2 = 0.04

It means that in a sample of size 81 light bulbs, we would observe a sample mean of 300 hours or more away from 7463 about 4% of the time by chance alone.

c) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × σ/√n

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, z score for 95% confidence level is 1.96

Margin of error = 1.96 × 1080/√81

= 235.2

Confidence interval = 7163 ± 23.2

a) Since alpha, 0.05 > than the p value, 0.04, then we would reject the null hypothesis. Therefore, at a 5% level of significance, there is evidence that the mean life is different from 7463 hours

Comparing the results of a and c, it is true that the population mean life is not 7463 hours.

4 0
3 years ago
2. The Fredericks family goes out for dinner, and the price of the meal is $60. The sales tax
Phoenix [80]
$60 x (100%+15%+7%) = $60 x 122% = $73.20
7 0
3 years ago
Lawrence performs a survey to determine the average number of minutes of exercise seventh-grade athletes get in one week. He get
RoseWind [281]

Answer:

1) Is more representative

Step-by-step explanation:

The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability

This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.

If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.

5 0
3 years ago
Read 2 more answers
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