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kondor19780726 [428]
3 years ago
12

Find the value of the expression 1/2x + 3 if x = 12

Mathematics
2 answers:
gregori [183]3 years ago
8 0

Step-by-step explanation:

please mark me as brainlest

liq [111]3 years ago
6 0
1/2(12) + 3 = 9

Change 1/2 into a decimal, multiply by 12, and then add 3
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Dell is a salesperson he sold a snowboard for $117 and earn 14% commission how much commission did Dell earn​
Lelu [443]

Answer:

Dell made $16.38

Step-by-step explanation:

4 0
3 years ago
CAN SOMEBODY SOLVE THIS AND GIVE ME A STEP-BY-STEP?<br> -40 = -5p
andrey2020 [161]

Answer:

p = 8

Step-by-step explanation:

-5p = -40 so 40/5 = 8

8 0
3 years ago
Use the given data to find the 95% confidence interval estimate of the population mean μ. Assume that the population has a norma
loris [4]

Answer: For 95% Confidence Interval:

Upper Limit = 110.2

Lower Limit = 97.8

95% Confidence Interval = [97.8, 110.2]

Step-by-step explanation:

Given that,

Mean(M) = 104

Standard Deviation(SD) = 10

Sample Size(n) = 10

Formula for calculating 95% Confidence Interval are as follows:

Standard error(SE) =\frac{SD}{\sqrt{n} }

                         = \frac{10}{\sqrt{10} }

                         = 3.164

⇒ M ± Z_{0.95} × SE

= 104 ± (1.96)(3.164)

= 104 ± 6.20

∴ Upper Limit = 104 + 6.20 = 110.2

   Lower Limit = 104 - 6.20 = 97.8

So,

95% Confidence Interval = [97.8, 110.2]

6 0
3 years ago
Haruka hiked several kilometers in the morning. She only hiked 6 kilometers in the afternoon, which was 25 percent less she had
Allisa [31]
6km is 3/4 of her morning speed so 6/(3/4)=6×4/3=8km. Total distance=6+8=14km.
5 0
4 years ago
Read 2 more answers
What is the solution to the system of equations? {10x^2 - y = 48 2y = 16x^2 + 48
julsineya [31]
A graph shows the solutions to be
  (x, y) = (-6, 312) or (6, 312)


To solve algebraically, remove a factor of 2 from the second equation and use that expression to substitute for y in the first equation.
  10x^2 -(8x^2 +24) = 48
  2x^2 = 72
  x = ±√(72/2) = ±6
  y = 8*36 +24 = 312

6 0
3 years ago
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