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s344n2d4d5 [400]

2 years ago

12

What is the solution to 5(3-x)?

1 answer:

Anit [1.1K]2 years ago

8 0

The answer is 15 - 5 x

Multiply 5 by 3 ( 15 )
Multiply 5 by -x ( -5x )

I hope this helps and please mark brainliest!

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PLS HELP ILL GIVE U BRAINLIEST

vekshin1

The answer to this question would be the second option “10m + 5 > 100” because 10 a month so (10m) then he purchases 5 extra per month so + 5 then his total collection is MORE so greater than 100 games so > 100 so the outcome is “10m + 5 > 100”

5 0

2 years ago

Read 2 more answers

umka2103 [35]

Answer:

8.48%

Step-by-step explanation:

Bill's weight decreases from 64.8 kg to 59.3 kg.

We need to find the percentage decrease in Bill's weight. It can be calculated as follows :

$\%=\dfrac{64.8 -59.3 }{64.8 }\times 100\\\\=8.48\%$

So, the required decrease in percentage is 8.48%.

3 0

2 years ago

What algebraic expression represents the phase “two times the quantity of a number minus 12”

MakcuM [25]

Answer:

2(n-12)

Step-by-step explanation:

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3 years ago

A polygon is reflected across a line. The angles in the original figure are compared to the corresponding angles in the image to

irina [24]

None. <span>They will always be congruent.</span>

5 0

3 years ago

Read 2 more answers

A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers

Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

$\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57$

Using normal distribution table, $z_{0.005$ is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

$e= z_{0.005}\frac{\sigma}{\sqrt{n} } = 2.57*\frac{3900}{\sqrt{100} } =1002.3$

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0

3 years ago