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lubasha [3.4K]
2 years ago
10

1. Which of the following is an example of acceleration? A. 10m/s B. 64km/h/min C. 50 cm2/s2 D. 46 km/h​

Mathematics
1 answer:
iogann1982 [59]2 years ago
3 0

Answer:

i beleive c for the proper answer form is m/s 2 c is the only one to show to square anything. if that's wrong the a but there is no square symbol.

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Given: 6(2x−2)≥3x+2(x+1)<br> Prove: x≥2
lyudmila [28]
I don’t know to be honest
8 0
3 years ago
Read 2 more answers
Please help me i know this might take a while but plz my teacher be buggin <br> Due today
trasher [3.6K]

Answer:

5) Part A: The rate of change, also known as the slope, is the "0.6x" part of the equation. It means that for every week the puppy is alive, x, that number is multiplied by 0.6 pounds.

Part B: The y-intercept is 4 pounds, which represents a puppy being 4 pounds at birth.

Part C: y = 0.6x + 4

y = 0.6(8)+4

y = 8.8

At 8 weeks, a puppy is estimated to be 8.8 pounds.

Part D: y = 0.6x + 4

22 = 0.6 + 4

-4              -4

18 = 0.6x

18/0.6 = 0.6x/0.6

30 = x

It would take a puppy 30 weeks to reach 22 pounds.

8 0
2 years ago
Helppppppppppppppppppppp
vitfil [10]
41 hours of class I guess so
5 0
2 years ago
Convert the following repeating decimal to a fraction in simplest form .31 with the 1 repeating
bekas [8.4K]

Answer:

14/45

Step-by-step explanation:

So we have the fraction:

0.3\bar1=0.31111...

We can do this algebraically. Follow to following steps:

Let's let this number equal to n. Thus:

0.31111...=n

Since there is only 1 digit repeating, let's multiply everything by 10. So:

3.1111...=10n

Now, subtract n from both sides:

3.1111-n=10n-n

On the left, substitute the number for n. On the right, combine like terms:

3.1111...-0.31111...=9n

All of the 1s will cancel. So:

3.1-0.3=9n

Subtract:

2.8=9n

Divide both sides by 9:

n=2.8/9

Remove the decimal by multiplying both sides by 10:

n=28/90

Reduce:

n=14/45

And we're done!

Use a calculator to check:

14/45\stackrel{\checkmark}{=}0.31111...

7 0
3 years ago
The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

5 0
3 years ago
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