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viktelen [127]
3 years ago
6

I need to fill this out for notes .

Mathematics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

not an answer but how do you attach images because it would be more useful

Step-by-step explanation:

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(t-4)(3t+1)(t+2)=0 hats the answer
trasher [3.6K]
(3t² +t -12t - 4)(t+2)=0
(3t³ - 5t² -26t -8) =0
t = -2
7 0
3 years ago
Read 2 more answers
Scores on an aptitude test are distributed with a mean of 220 and a standard deviation of 30. The shape of the distribution is u
Evgen [1.6K]

Answer:

P(215<X<225) = 0.7620

Step-by-step explanation:

The shape of the distribution is unknow, however, the shape of the sampling distributions of the sample mean is approximately normal due to the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 220, \sigma = 30,n = 50, s = \frac{30}{\sqrt{50}} = 4.2426

P(215<X<225)

This is the pvalue of Z when X = 225 subtracted by the pvalue of Z when X = 215. So

X = 225

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{225 - 220}{4.2426}

Z = 1.18

Z = 1.18 has a pvalue of 0.8810

X = 215

Z = \frac{X - \mu}{s}

Z = \frac{215 - 220}{4.2426}

Z = -1.18

Z = -1.18 has a pvalue of 0.1190

0.8810 - 0.1190 = 0.7620

P(215<X<225) = 0.7620

6 0
3 years ago
Q1: Which of the displays would be easiest to use to find the mean of a set of data?
elena-14-01-66 [18.8K]

Answer:

I would say for the first one a dot plot and the second one a box plot. I hope I'm right! Sorry if I'm wrong.

Step-by-step explanation:

because a box plot shows you the median

4 0
3 years ago
I’ll give brainliest , 7th grade work !
NARA [144]

Answer:

5.90551

Step-by-step explanation:

7 0
3 years ago
See image attached below keeeeeeeeeeeeeeeeeeeed
emmasim [6.3K]

Answer:

.13%

68.26%

2.28%

47.72%

49.87%

34.13%

Step-by-step explanation:

1.) standardize by subtracting the mean and dividng by the standard deviation

(625-1000)/125= -3

go to a ztable to get .0013 or (1-.9987)

this is equal to .13%

2.)

875<x<1125

standardize both seperately and subtract them

(875-1000)/125= -1  whose probability is .1587 (or 1-.8413)

(1125-1000)/125= 1  whose probability is .8413

.8413-.1587= 68.26%

3.)

Find the probability that someone paid less than 1250 and take its compliment

(1250-1000)/125= 2 which has a probability of .9772

take its compliment (1-.9772)= .0228= 2.28%

4.)

Same process as question 2

(750-1000)/125= -2 which has a probability of (1-.9772) = .0228

(1000-1000)/125= 0 which has a probability of .5

.5-.0228= .4772= 47.72%

5.) same deal as the previous question

(625-1000)/125= -3 which has a probability of (1-.9987)= .0013

(1000-1000)/125= 0 which has a probability of .5

.5-.0013= .4987= 49.87%

6.)same deal the previous question

(875-1000)/125= -1 which has a probability of (1-.8413)=.1587

(1000-1000)/125= 0 which has a probability of .5

.5-.1587= .3413= 34.13%

4 0
3 years ago
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