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Gala2k [10]
3 years ago
11

Ik this is easy but i forgot how to do it help​

Mathematics
1 answer:
REY [17]3 years ago
8 0

Answer:

7x

Step-by-step explanation:

4+3= 7

=7x

(since they both have the variable "x" we can combine them both)

have a good day! <3

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Carl spent $140 in six days. He spent $20 on the first day. He spent the same amount of money during the next five days Write an
grin007 [14]

Answer:

answer: he spent $120

Step-by-step explanation:

20×5=100

100+ the initial $20 spent = $120

8 0
2 years ago
Solve 2(3-a)=18 for a
Alinara [238K]

Answer:

a=-6

Step-by-step explanation:

2(3-a)=18

3-a=18/2

3-a=9

a=3-9

a=-6

7 0
3 years ago
Read 2 more answers
TO get a certain shade of green paint, Melissa decides to mix 3 gallons of blue paint with 2 gallons of yellow paint. How much b
Harlamova29_29 [7]
You have a fraction of 3/2. 3/2=1.5
Therefore, she needs 1.5 gallons of blue paint per 1 gallon of yellow paint.
7 0
3 years ago
Help plz i’ll give extra points f
lys-0071 [83]

Answer: First Option | 2 x 10^1

Reason: After doing 6.4 divided by 3.2, you get 2, and finally you subtract 10³ by 10². And after that subtraction, you get 10^1, Therefore the answer is the first option.

<em>I hope this helps, and Happy Holidays! :)</em>

3 0
2 years ago
Question 7 of 10
andriy [413]

Answer:

C. n = 90; p = 0.8

Step-by-step explanation:

According to the Central Limit Theorem, the distribution of the sample means will be approximately normally distributed when the sample size, 'n', is equal to or larger than 30, and the shape of sample distribution of sample proportions with a population proportion, 'p' is normal IF n·p ≥ 10 and n·(1 - p) ≥ 10

Analyzing  the given options, we have;

A. n = 45, p = 0.8

∴ n·p = 45 × 0.8 = 36 > 10

n·(1 - p) = 45 × (1 - 0.8) = 9 < 10

Given that for n = 45, p = 0.8, n·(1 - p) = 9 < 10, a normal distribution can not be used to approximate the sampling distribution

B. n = 90, p = 0.9

∴ n·p = 90 × 0.9 = 81 > 10

n·(1 - p) = 90 × (1 - 0.9) = 9 < 10

Given that for n = 90, p = 0.9, n·(1 - p) = 9  < 10, a normal distribution can not be used to approximate the sampling distribution

C. n = 90, p = 0.8

∴ n·p = 90 × 0.8 = 72 > 10

n·(1 - p) = 90 × (1 - 0.8) = 18 > 10

Given that for n = 90, p = 0.9, n·(1 - p) = 18 > 10, a normal distribution can be used to approximate the sampling distribution

D. n = 45, p = 0.9

∴ n·p = 45 × 0.9 = 40.5 > 10

n·(1 - p) = 45 × (1 - 0.9) = 4.5 < 10

Given that for n = 45, p = 0.9, n·(1 - p) = 4.5 < 10, a normal distribution can not be used to approximate the sampling distribution

A sampling distribution Normal Curve

45 × (1 - 0.8) = 9

90 × (1 - 0.9) = 9

90 × (1 - 0.8) = 18

45 × (1 - 0.9) = 4.5

Now we will investigate the shape of the sampling distribution of sample means. When we were discussing the sampling distribution of sample proportions, we said that this distribution is approximately normal if np ≥ 10 and n(1 – p) ≥ 10. In other words

Therefore;

A normal curve can be used to approximate the sampling distribution of only option C. n = 90; p = 0.8

3 0
2 years ago
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