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PIT_PIT [208]
3 years ago
15

Sum of first 1000 consecutive odd counting numbers

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

100,000,000

Step-by-step explanation:

did some research to solve this one, i recommend looking more into to it to ensure that this is the correct answer ;)

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57.42
Fofino [41]

Hi there!

\large\boxed{a = 6 m/s^{2}}

Use the kinematic equation to solve for acceleration:

a = \frac{v_{f}-v_{i}}{t}

Where:

vf = final velocity

vi = initial velocity

t = time

Plug in the given values:

a = \frac{65-35}{5}\\\\a = 30 / 5  \\\\=6 m/s^{2}

7 0
3 years ago
a school wants to make a new playground by cleaning up and abandoned lot that is shaped like a rectangle. they give the job of p
svetoff [14.1K]
 <span>3/8 is left is the correct answer if multiple choice

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5 0
3 years ago
Kelly spends 3/8 of her free time playing soccer. what percent of her free time was spent playing soccer?
lozanna [386]
I think it's 37.5 . I'm pretty sure cause I think you divide 3/8 . Then you go 2 to the right for your answer with the decimal. Not sure but . I think I'm right.
7 0
3 years ago
Please help It’s almost due
Tems11 [23]

Answer:

10 people

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

                                  T.S.  = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

                               =  -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0
3 years ago
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