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3 years ago

14

When Joseph’s first starts working at a grocery store, his hourly rate is $10. For each year he works at the grocery store, his

hourly rate increases by $0.50.

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

R(t) = 10 + 0.50t

Step-by-step explanation:

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I need help dont get it

Mariulka [41]

Answer:

a.) y= 25 + 5x

b.) 45$

Step-by-step explanation:

25$ from the start and adds 5 for extra

y= 25+5(4)

y=25+20

y=45$

7 0

3 years ago

Evaluate the expression a2+6ab for a=−3 and b=4.

inn [45]

First you plug in the variables:

(-3)(2)+(6)(-3)(4)

Multiply:
-6+(-18)(4)
-6+-72

Add and your answer comes out to be:
-78

5 0

3 years ago

How many people own a SUV?<br> A. 4<br> B. 42<br> C. 420<br> D. 4,200

MaRussiya [10]

For this, no math is even needed! The question says 1,000 people were interviewed. 42% percent of that 1,000 owns an SUV. If it's a percent, it is 42/100, right? Now we have to make that 42/100 out of 1000. We add a zero to the 100 on the bottom (42/1000). We do the same thing on the top. Add a zero! 420.

SECOND POSSIBLE SOLUTION
Solve this with logic! 42% is nearly 50%. Half of 1,000 is 500. What is closest to 500? 420! This solution is even easier.

Hope this helps you!

5 0

4 years ago

Pls help I am giving brainliest help me ASaP

Goryan [66]

Answer:

4

Step-by-step explanation:

4 because 4^2 = 16.

5 0

3 years ago

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

4 years ago