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MA_775_DIABLO [31]
4 years ago
5

Help asap???

Mathematics
1 answer:
USPshnik [31]4 years ago
5 0

Answer:

the probability of picking a black clothing is 12/30 (6/15), and the probability of picking a pair of pants is 17/30

Step-by-step explanation:

First we have to figure out how much will it be out of, to do that you have add the amount of clothing there is (in this case 30).

Determine how many pairs of black clothing:

you must add up the amount of black clothes so 5+7=12, so there are 12 black clothing and 30 in total so the chance is 12/30 if you want it reduced it could be 6/15.

determine how many pairs of pants:

Now it's the same as last step add up all the pairs of pants the question has so 7+10=17. so there are 17 pants out of 30 pieces of clothes so 17/30

(sorry if this isnt a good explanation, I tried)

You might be interested in
Which is greater |60| or |-60|
Aneli [31]

Answer:

They are both equal

Step-by-step explanation:

60 stays 60, and -60 converts to 60, so they are both 60

60=60

Hope it helped!~❤️

6 0
3 years ago
Read 2 more answers
Priyas cat weighs 5 and 1/2 lbs and her dog weighs 8 and 1/4 lbs.
sertanlavr [38]

Answer:

the answer is 13 and 3/4 and the dog is 2 and 3/4 im pretty sure

Step-by-step explanation:

you add for the fisrt one than subtract for the second one

4 0
4 years ago
Read 2 more answers
0.85 0.88 0.88 1.06 1.09 1.12 1.29 1.31 1.42 1.49 1.59 1.62 1.65 1.71 1.76 1.83 Assume that the distribution of coating thicknes
iren2701 [21]

Answer:

μ = 1.3468

Step-by-step explanation:

Given data:

Observations

0.85

0.88

0.88

1.06

1.09

1.12

1.29

1.31

1.42

1.49

1.59

1.62

1.65

1.71

1.76

1.83

sum of the observations = 21.55

now,

the total number of the given data, n = 16

now, the mean is given as:

Mean = (Sum of all the observation) / (Total number of observations)

on substituting the values in the above formula, we get

Mean = 21.55 / 16

or

Mean = 1.3468

Since, it is given that the data follows normal distribution.

thus, the normal mean can be used as the point of estimator (μ)

i.e μ = 1.3468

3 0
3 years ago
If he is correct, what is the probability that the mean of a sample of 68 computers would differ from the population mean by les
elena-14-01-66 [18.8K]

Complete Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad

Answer:

P(-1.72

Step-by-step explanation:

From the question we are told that:

Population mean \mu=91

Sample Mean \=x =2.08

Standard Deviation \sigma=10

Sample size n=68

Generally the Probability that The  sample mean  would differ from the population mean

P(|\=x-\mu|<2.08)

From Table

P(|\=x-\mu|

T Test

Z=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }

Z=\frac{2.08}{\frac{10}{\sqrt{68} } }

Z=1.72

P(|\=x-\mu|

P(-1.72

Therefore From Table

P(-1.72

5 0
3 years ago
Express each percent as a fraction in simplest form.<br>a. 85%<br>b. 5 72%<br>c. 12.55%​
iVinArrow [24]

Answer:

(a) 17/20 b.5/18/25 c. 1.255

8 0
3 years ago
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