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masha68 [24]
3 years ago
11

GIVING BRAINLIST PLS HELP ASAP !!!!!

Mathematics
1 answer:
matrenka [14]3 years ago
8 0

Answers and work shown in attachment! :)

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Solve for the missing number, founding to three decimal places when necessary<br> b. 0.76 =<br> 1425
boyakko [2]

Answer:

1.425

Step-by-step explanation:

b.0.76

0.76÷1000=0.076

l425÷1000=1.425

5 0
3 years ago
Help please :( points !
mars1129 [50]
Answer: 2/5

Explanation: 10+8+12 = 30 12/30 = .4 .4 = 2/5
4 0
3 years ago
Read 2 more answers
Solve the equation<br><br> 1.6=7.6-5(k+1.1)
zepelin [54]
7.6 - 5(k + 1.1) = 1.6

7.6 - 5(k + 1.1) - 7.6 = 1.6 - 7.6

-5(k +1.1) = -6

-5(k + 1.1) -6
-------------- = ---------
-5 -5

k + 1.1 = 6
-----
5

k + 1.1 - 1.1 = 6
----- - 1.1
5

k = 0.1
8 0
3 years ago
NEED HELP ASAP!!!!!!!!!!
WITCHER [35]

Answer:

the question is missing

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A survey found that​ women's heights are normally distributed with mean 63.2 in. and standard deviation 2.4 in. The survey also
Effectus [21]

Answer: 99.51%

Step-by-step explanation:

Given : A survey found that​ women's heights are normally distributed.

Population mean : \mu =63.2  \text{ inches}

Standard deviation: \sigma= 2.4\text{ inches}

Minimum height = 4ft. 9 in.=4\times12+9\text{ in.}=57\text{ in.}

Maximum height = 6ft. 2 in.=6\times12+2\text{ in.}=74\text{ in.}

Let x be the random variable that represent the women's height.

z-score : z=\dfrac{x-\mu}{\sigma}

For x=57, we have

z=\dfrac{57-63.2}{2.4}\approx-2.58

For x=74, we have

z=\dfrac{74-63.2}{2.4}\approx4.5

Now, by using the standard normal distribution table, we have

The probability of women meeting the height requirement :-

P(-2.58

Hence, the percentage of women meeting the height requirement = 99.51%

8 0
3 years ago
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