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sergejj [24]
3 years ago
12

Simplify 0.05x + 1.02x. 0.17x 1.07x 1.7

Mathematics
2 answers:
Kitty [74]3 years ago
7 0
0.05x+1.02x=(0.05+1.02)x=\boxed{1.07x}
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

1.07x

I think i got it right..wait lemme check........yep i gots it right..hope it helped

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Find the factorization of each of the following number:<br> A.18 B.24 C.38 D.81
const2013 [10]

Answer:

A: (3^2)*2

B: (2^3)*3

C: 2*19

D: (3^2)*9

Step-by-step explanation:

We have:

A) 18= 3*3*2= (3^2)*2

B) 24=2*2*2*3=(2^3)*3

C) 38 = 2*19

D) 81 =3*3*9=(3^2)*9

6 0
3 years ago
Find the slop of the line​
Goshia [24]

Answer:

5/4 is your slope.

Step-by-step explanation:

5-0= 5

4-0= 4

slope is chang in y over change in x. Your first point is (0,0) and you second is (4,5).

6 0
3 years ago
&lt;1 and &lt;2 are supplementary angles. The measure of &lt;1 is 87º. The measure of &lt;2 is 15x + 18º.
Kay [80]

Answer: x= 5

Step-by-step explanation: If angles 1 and 2 are supplementary, that means they add up to 180. So if angle 1 is 87, angle 2 has to be 180-87=93. Then the equation would be 15x+18=93. Subtract 18 on both sides. 15x=75. Divide by 15 and your answer is 5.

6 0
3 years ago
In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was nor
Alenkinab [10]

Answer:

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P(Xbar>102.5).

P(Xbar>102.5)=P(Z>1.33)

P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)

P(Xbar>102.5)=0.5-0.4082

P(Xbar>102.5)=0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.

5 0
3 years ago
I need help with one more and can you explain how it works too thank you
notka56 [123]
The answer to this question is 5.2

8 0
3 years ago
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