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frez [133]
3 years ago
11

Do yall know what 81 ➗ 918 = if yall know ill give brainleist

Mathematics
1 answer:
Ksju [112]3 years ago
6 0
0.088235294117647 (20 characters thing)
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A researcher is interested in estimating the mean weight of a semi trailer truck to determine the potential load capacity. She t
Kaylis [27]

Answer: 95% confidence interval = 20,000 ± 2.12\times\frac{1500}{\sqrt{17} }

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

Step-by-step explanation:

Given :

Sample size(n) = 17

Sample mean = 20000

Sample standard deviation = 1,500

5% confidence

∴ \frac{\alpha}{2} = 0.025

Degree of freedom (d_{f}) = n-1 = 16

∵ Critical value at ( 0.025 , 16 ) = 2.12

∴ 95% confidence interval = mean ± Z_{c}\times\frac{\sigma}{\sqrt{n} }

Critical value  at 95% confidence interval = 20,000 ± 2.12\times\frac{1500}{\sqrt{17} }

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

3 0
3 years ago
Answer if you want, 10 points
rjkz [21]

Answer:

hi

Step-by-step explanation:

for example

\frac{8}{4} = 2 \\  \\ 1 \times \frac{4}{5}  =  \frac{9}{5}

in the second one you should multiple 5 and 1 then add 4

8 0
3 years ago
X^2 + y^2=40 and y=x-4 solve this simultaneous equation
inn [45]
Write x-4 instead of y.
x²+(x-4)²=40
x²+x²-8x+16=40
2x²-8x-24=0
x²-4x-12=0
(x-6)(x+2)=0
x=6 or x=-2

for x=6, y=6-4 = 2
for x=-2, y=-2-4 = -6
3 0
3 years ago
Noah applies two transformations to WXYZ so that the final vertices of the transformed
krok68 [10]

Answer:

ok

Step-by-step explanation:

4 0
3 years ago
If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many
kotykmax [81]

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of  M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the  first successful trial, then the probability of having k is given by

Pr(X=k) = (1-p)^{k-1}p  

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

4 0
3 years ago
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