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harina [27]
2 years ago
13

Name 4 fractions that are equivalent to 9/12. Show your work

Mathematics
2 answers:
Alenkasestr [34]2 years ago
7 0

Answer:

3/4, 18/24, 27/36, 36/48

Step-by-step explanation:

9 and 12 can both be divided by 3 to equal 3/4

9/12 times 2 is 18/24

9/12 times 3 is 27/36

9/12 times 4 is 36/48

hope this helps:)

Andrei [34K]2 years ago
5 0

Answer:

3/4

6/8

12/15

15/18

Step-by-step explanation:

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What’s the Value for k?
xeze [42]

Answer:

Step-by-step explanation:

The sum of the angles on a straight line is 180 degrees. Therefore,

Angle XYZ + angle MYZ = 180

Angle XYZ + 115 = 180

Angle XYZ = 180 - 115 = 65 degrees

The sum of the angles in a triangle is 180 degrees. It means that

Angle XZY + angle YXZ + angle MYZ = 180

Therefore,

4k + 5 + 6k + 10 + 65 = 180

4k + 6k + 5 + 10 + 65 = 180

10k + 80 = 180

10k = 180 - 80 = 100

Dividing the left hand side and the right hand side of the equation by 10, it becomes

10k/10 = 100/10

k = 10

5 0
3 years ago
A science teacher ordered $1848 worth of microscopes and dissection kits for the new science lab. A total of 30 microscopes and
andrezito [222]

Answer:

Microscopes cost $44 and Dissection Kits cost $11

Step-by-step explanation:

Using the information given, we can make the system of equations:

30x+48y=1848

x represents the price of microscopes and y represents the price of dissection kits.

We also know that microscopes will be 4x the price of dissection kits. Therefor:

x=4y

We can plug these two equations together to find the answer.

30(4y)+48y=1848

120y+48y=1848

168y=1848

y=11

Using the y-value, can find the x-value.

x=4(11)

x=44

4 0
3 years ago
PLEASE HELP (20 POINTS)
Paul [167]

Answer:

option b

Step-by-step explanation:

you can't take out anything, not an x or a 2

6 0
3 years ago
can someone solve these two equations 2/8=3/r and the next problem is 2/5=9m/8 these are two different problems
Drupady [299]
(1) 3/4=r

(2) 16/45=m
7 0
3 years ago
A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after
lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                         P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of persons who were still employed primarily as engineers  = \frac{111}{200} = 0.555

           n = sample of graduates = 200

           p = population proportion of engineers

<em>Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                 significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } , 0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } } ]

 = [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0
3 years ago
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