-5x^5+35x^4-30x^3 can someone help me pls

Complete the following table for residuals for the linear function f(x) = 138.9x − 218.76. (Round to the hundredths place)

OLEGan [10]

Answer:

<h3>i believe it is 65 if it isnt then idk</h3>

Step-by-step explanation:

6 0

3 years ago

Which expression is equivalent to 2x^2+2x-4/2x^2-4x+2

lord [1]

You need to factor the numerator and denominator...

(2x^2+4x-2x-4)/(2x^2-2x-2x+2)

(2x(x+2)-2(x+2))/(2x(x-1)-2(x-1))

((2x-2)(x+2))/((2x-2)(x-1)) so the (2x-2)s cancel out leaving

(x+2)/(x-1)

4 0

3 years ago

Read 2 more answers

Can you name some math terms starting with the letter "O" and give a short description of them?

likoan [24]

Obtuse angle

Operation

Order of operation

4 0

3 years ago

Read 2 more answers

Write the recursive formula then determine the next term of 3,10,24,52

Schach [20]

Answer:

f[1] = 3
f[n] = 2·f[n-1] +4
108

Step-by-step explanation:

We observe that first differences of the given numbers are ...

10 -3 = 7

24 -10 = 14

52 -24 = 28

That is, each difference is 2× the previous one. This suggests an exponential relation that has a base of 2.

We notice that doubling a term doesn't give the next term, but gives a value that is 4 less than the next term. So, we can get the next term by doubling the previous one and adding 4.

Then our recursive relation is ...

f[1] = 3 . . . . the first term

f[n] = 2×f[n-1] +4 . . . . double the previous term and add 4

The next term is 2·52 +4 = 108.

7 0

4 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%.

Let $\mu$ = true average percentage of organic matter in such soil

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

Alternate Hypothesis, $H_A$ : $\mu \neq$ 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 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 amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 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.

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.

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

3 0

3 years ago