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2 years ago

10

A mixture of green paint contains 9 pounds of blue paint and 12 pounds of yellow paint to make the same shade of green paint usi

ng 21 pounds of blue paint how much yellow paint would you need

1 answer:

kari74 [83]2 years ago

4 0

I’m not too sure, I tried dividing 21 by 9 which then I got 2.3333 so then I multiplied that and 12 and got basically 28

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When Andrei surveyed 36 random seventh-grade students in the lunchroom, he found that 7 out of 9 would like to try or have

Alex73 [517]

Answer:

About 630 seventh-grade students would like to snorkel

Step-by-step explanation:

if you do

7/9 x 90 equals 630/810 total students

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3 years ago

Math inequality<br><br> y+ 3 < 10

wariber [46]

Answer:

y <7

Step-by-step explanation:

Move all terms not containing y to the right side of the inequality.

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2 years ago

Read 2 more answers

Verify the identity 6cos^2(x) -3 = 3 - 6sin^2(x)

Lorico [155]

Answer:

It is proved that $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$ .

Step-by-step explanation:

We already have the identity of x as $\sin ^{2}x + \cos ^{2}x = 1$ .......... (1) .

So, from equation (1) we can write that

$\cos ^{2} x = 1 - \sin ^{2} x$

⇒ $6\cos ^{2} x = 6 - 6 \sin ^{2} x$

⇒ $6\cos ^{2} x -3 = 6 - 6 \sin ^{2}x -3$

⇒ $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$

Hence, it is proved that $6\cos ^{2}x -3 = 3 - 6\sin ^{2} x$ . (Answer)

5 0

3 years ago

Find the value of the missing variables which are M, A, and X

Mars2501 [29]

Answer:

m=5 , a=3 x= 3 you get that by looking at the other numbers that they provided.

3 0

2 years ago

Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0

2 years ago