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3 years ago

PLEASE ANSWER + BRAINLIEST!!!

Mathematics

2 answers:

Step2247 [10]3 years ago

8 0

The second expression

konstantin123 [22]3 years ago

7 0

Here is the correct answer of the given polynomial above. The answer is [(-3x^2)+(-8x^2)]+4x+[9+5]. The expression that can be used to find the sum of the polynomials given above is [(-3x^2)+(-8x^2)]+4x+[9+5].

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What is the 14th term of the sequence?

fomenos

Answer:

D. 16,384

Step-by-step explanation:

The given sequence (including "6") is neither arithmetic nor geometric. The 14th term cannot be predicted. (It can be anything you like.)

Based on the answer choices, we assume the intended sequence is the geometric sequence ...

2, 4, 8, 16, 32, ...

which has formula ...

an = 2^n

Then the 14th term is ...

a14 = 2^14 = 16,384

3 0

3 years ago

Write the equation of a line whose graph has the same slope as

Sveta_85 [38]

Answer:

y= 1/3x -3

Step-by-step explanation:

For the y-int for 8x-2y=6 is (0, -3).

For the slope of 3x-9y=9 is 1/3.

3 0

3 years ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0

3 years ago

Factor the expression <br> -3h -9

valina [46]

Answer:

- 3(h + 3)

Step-by-step explanation:

- 3h - 9 ← factor out - 3 from each term

= - 3(h + 3)

7 0

3 years ago

Pls help asap I’ll mark BRAINLIEST DONT answer if your not sure, this affects my whole grade

Sidana [21]

Answer:

2.5

Step-by-step explanation:

Add up the 5s of the left side first and you will get 25. Then you add up the 5s on the otherside and you will get 10. Subtract the 10 from the 25 to get 15. Finally, divide 15 by 6 to get the weight of each triangle, 2.5

1: 5 x 5 = 25

2: 5 x 2 = 10

3: 25 - 10 = 15

4: 15/6 = 2.5

The weight of each triangle is 2.5

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6 0

3 years ago

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