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lara [203]
3 years ago
12

Please help me with this question ​

Mathematics
1 answer:
WINSTONCH [101]3 years ago
5 0

Answer:

it is not linear

Step-by-step explanation:

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What is 4r+9r-11r+7r please im not trying to fail
kaheart [24]

Answer:

9r

Step-by-step explanation:

combine like terms in PEMDAS order. Add 4, 9, and 7, then, subtract 11.

5 0
3 years ago
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A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get
julsineya [31]

I have an expression

\sigma = \sqrt{p(1-p)/n}

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

E( (b-p)^2 ) =  (-p)^2(1-p) + (1-p)^2p = p(1-p)

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

\sigma = \sqrt{p(1-p)/n}

Plugging in the numbers,

\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

7 0
3 years ago
Solve for a = ____.<br><br> 1. 6<br> 2. 4<br> 3. 9
garik1379 [7]
The answer is 3, or 9
8 0
3 years ago
A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average i
strojnjashka [21]

The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

  • Population mean = \mu = $150
  • Population standard deviation = \sigma = $30.20
  • Sample mean = \overline{x} = $160
  • Sample size = n = 40 > 30
  • Level of significance = \alpha = 2.5% = 0.025
  • We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

  • Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: H_0: \mu_0 \leq \mu = 150
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically H_1: \mu_0 > \mu = 150

where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

3 0
2 years ago
Read 2 more answers
Please I will give brainliest​
atroni [7]

Answer:

I think its D because i did the math for it but tell me if im wrong.

Step-by-step explanation:

I hope you  have a nice day

8 0
2 years ago
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