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3 years ago

Order the following numbers from least to greatest.

Mathematics

2 answers:

Vladimir [108]3 years ago

8 0

Answer:

-2,1,4,5

Step-by-step explanation:

Volgvan3 years ago

5 0

-2, 1, 4, 5
hope that helps!

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A cake shop can make 240 boxes of cakes for 8 days. How many cakes can the shop make in 12 days?

denpristay [2]

Answer:

<h2>360 cakes</h2>

Step-by-step explanation:

<h2>soln:</h2><h2>onecake =240\8=30</h2><h2>then 12 cakes =30×12=360</h2>

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2 years ago

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Artyom0805 [142]

Answer:

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2 years ago

Gary learned that the value of his car depreciates by 15% percent per year. Which of the following functions best describes the

Sever21 [200]

For this case we have that the original value of the car is:
m dollars
For the following year we have the value is:
((100-15) / (100)) m
Rewriting we have:
((85) / (100)) m
0.85m
Answer:
the value of his car the year after the car is worth m dollars is:
B.f (m) = 0.85m

5 0

3 years ago

F(x)=3^2-2and g(x) =4x+2 what is the value of (f+g)(2)

morpeh [17]

Step-by-step explanation:

$f(x) = {3}^{2} - 2 = 9 - 2 = 7 \\ g(x) = 4x + 2 \\ \therefore \: (f + g)(x) = 7 + 4x + 2 \\ \therefore \: (f + g)(x) = 4x + 9 \\ \therefore \: (f + g)(2) = 4 \times 2 + 9 \\ \therefore \: (f + g)(2) = 8 + 9 \\ \red{\boxed{ \bold{ \therefore \: (f + g)(2) = 17}}}$

4 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

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

<em>Let </em> $\mu$ <em> = true average percentage of organic matter in such soil</em>

SO, <u>Null Hypothesis</u>, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

<u>Alternate Hypothesis</u>, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is <u>One-sample t test statistics</u> 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 amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, <u><em>test statistics</em></u> = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

<u></u>

<u>Now, P-value of the test statistics is given by;</u>

P-value = P( $t_2_9$ > -1.759) = <u>0.046</u> or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago