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11111nata11111 [884]
3 years ago
11

What are the factors of the expression below? -2x+5x-3

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

-2x-15

Step-by-step explanation:

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N − 5 ≥− 2 please show your work to how u got your answer.
kondaur [170]

Answer:

n ≥ 3

Step-by-step explanation:

add (-5) to both sides of equations. Solve for n.

3 0
3 years ago
Read 2 more answers
One benefit of cluster sampling is that it Group of answer choices
luda_lava [24]

Answer:

a. is effective for use in school research.

Step-by-step explanation:

Cluster sampling is the sampling whereby we pick random few sample from the population to serve as the sample population called cluster. This cluster usually have uniform characteristics among them.

The main advantage of cluster sampling is that it is effective school research because of its cheapness and doesn't require much resources during the sampling process.

The other options are not correct because Cluster sampling have High sampling error and its biased sample.

3 0
3 years ago
PLS SOLVE THIS<br> I WILL MARK BRAINLIEST
Paraphin [41]

Answer:

AB = 25 units

Step-by-step explanation:

In right triangle ACB, CD\perp AB

Therefore, by geometric mean property:

CD^{2} ={AD\times DB}  \\  \\  {10}^{2}  = AD\times 5 \\  \\ 100 = 5AD \\  \\  AD =  \frac{100}{5}  \\  \\ AD = 20

AB = AD + DB

AB = 20 + 5

AB = 25 units

3 0
2 years ago
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0
3 years ago
In a market without environmental regulations, will the supply curve for a firm account for private costs, external costs, both,
bixtya [17]

Answer:

Step-by-step explanation:

It will only take into account private cost because it is allowed to emit pollution at a zero cost.

7 0
3 years ago
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