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2 years ago

Please explain how you got the answer.

Mathematics

2 answers:

photoshop1234 [79]2 years ago

8 0

Answer:

Step-by-step explanation:

First, simplify the equation they give you

125x+200≥1200 ---------> 125x≥1000---------->x≥8

So, x has to be greater than or equal to 8 which rules out b or d.

So the answers are narrowed down to a and c, the only difference is that the dot in A is filled in whereas the dot in c is not. In math, having a filled in circle means that value is included (≥ sign), so we will choose the answer a.

zubka84 [21]2 years ago

8 0

Anwser: A

step by step explanation:

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Cindy spent $15.25 on ingredients for a blueberry pie and $12.84 on ingredients for a cherry pie. Each slice of pie sells for $3

Brums [2.3K]

Answer:

61.07 dollars profit

Step-by-step explanation:

15.25 for blueberry pie = 24 slices (8 slices in a pie)

12.84 for cherry pie = 15 slices (8 slices in a pie)

24x3.50=84

15x3.50=52.5

2x12.84=25.68

3x15.25=45.75

84-45.75=34.25

52.5-25.68=26.82

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3 years ago

Read 2 more answers

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enyata [817]

Answer:well you see this is easy first you look at the 113 divide that by 4 multiply by 2 and then square root and you have your answer which would be d <4

Step-by-step explanation:

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3 years ago

The law f(t) = -t2 + 6t + 15 represents the number of kilometers of congestion, as a function of time, registered in a city. Wit

andrey2020 [161]

Time is -b/2a, which is 3.
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3 0

3 years ago

a field can be plowed by 8 machines in 6 hours. How many hours will it take 5 machines to plow the same field?

aniked [119]

Answer:

Step-by-step explanation:

If it's a riddle, then the answer is no time at all, simply because the field is already plowed.

8 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago