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ikadub [295]
2 years ago
15

To make grey paint, you need to mix ○ 2 cans of black paint ○ 3 cans of white paint If you have 5 cans of black paint, how many

cans of white paint do you need to make the grey paint? I NEED HELP ASAPPPPP
Mathematics
1 answer:
babymother [125]2 years ago
3 0
6.5 i believe. i think that’s the answer.
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We have 20 weeks left from school before the summer. How many days is left? Show me how you found the answer.
Troyanec [42]

Answer:

140 days

Step-by-step explanation:

there are 7 days in 1 week so just multiply 20 by 7 which is 140, to make it easier multiply 2 by 7 which is 14 then add the 0 soo 140.

Hope this helps :)

5 0
2 years ago
Read 2 more answers
What is the domain of the function graphed below?
asambeis [7]

Answer:

- 2 < x \leqslant 0 \: \: for \: x \leqslant 0

0 < x \leqslant 4 \:  and \: x \geqslant 7 \: \: for \: x \:  > 0

4 0
2 years ago
Rhonda deposited $3000 in an account in the Merrick National Bank, earning 4.2% interest, compounded annually. She made no depos
mafiozo [28]

Answer:

below

Step-by-step explanation:

that is the procedure above

3 0
3 years ago
Jane wants to estimate the proportion of students on her campus who eat cauliflower. after surveying 39 ​students, she finds 4 w
stellarik [79]

Let x be the number of students who eat cauliflower.

Therefore, x=4.

Let n be the total number of students surveyed.

Therefore, n=39

Thus, \hat p=\frac{4}{39} =0.10256

Now, for 90% confidence level, from the table we know that Z=1.645.

The formula for the interval range of proportion of students is :

p= \hat p\pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}}

Plugging in the values we get:

p=0.10256\pm 1.645\sqrt{\frac{0.10256(1-0.10256)}{39}}=0.10256\pm 0.04858=0.15114, 0.05398

Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).

7 0
3 years ago
Musical megabytes How much disk space
alexdok [17]

Given:

Consider the file sizes (in  megabytes) for 18 randomly selected files on  Gabriel's mp3 player are shown in the below figure.

To find:

The outliers in the  distribution.

Solution:

We have, the given data set

2.4, 2.7, 1.6 , 1.3 , 6.2, 1.3, 5.6, 1.1, 2.2, 1.9, 2.1, 4.4, 4.7, 3.0, 1.9, 2.5, 7.5, 5.0

Arrange the data in ascending order.

1.1, 1.3, 1.3, 1.6, 1.9, 1.9, 2.1, 2.2, 2.4, 2.5, 2.7, 3.0, 4.4, 4.7, 5.0, 5.6, 6.2, 7.5,

Divide the data in 4 equal parts.

(1.1, 1.3, 1.3, 1.6), 1.9, (1.9, 2.1, 2.2, 2.4),( 2.5, 2.7, 3.0, 4.4), 4.7, (5.0,5.6, 6.2, 7.5)

It is clear that,

Q_1=1.9

Q_3=4.7

IQR=Q_3-Q_1

IQR=4.7-1.9

IQR=2.8

Now,

Interval=[Q_1-1.5IQR,Q_3+1.5IQR]

Interval=[1.9-1.5(2.8),4.7+1.5(2.8)]

Interval=[-2.3,8.9]

All the data values lies in the interval [-2.3,8.9]. Therefore, the given data set have no outliers.

7 0
2 years ago
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