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ANTONII [103]
3 years ago
13

27 as 3 to a power.

Mathematics
1 answer:
dolphi86 [110]3 years ago
6 0
3^3 is equal to 3 * 3 *3, which is equal to 27
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-9 divided by 5/8 equals
DedPeter [7]
<span>-9 divided by 5/8 equals
-9 / (5/8) <---</span><span>Turn the second </span>fraction<span> upside down, then multiply.

= -9 * 8/5
= -72/5
= -14 2/5
= -14.4</span>
4 0
3 years ago
A sample of 11001100 computer chips revealed that 62b% of the chips fail in the first 10001000 hours of their use. The company's
STALIN [3.7K]

Answer:

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

z score = 1.35

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

Question; A sample of 1100 computer chips revealed that 62% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 60% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

Step-by-step explanation:

Given;

n=1100 represent the random sample taken

Null hypothesis: H0 = 0.60

Alternative hypothesis: Ha <> 0.62

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1100

po = Null hypothesized value = 0.60

p^ = Observed proportion = 0.62

Substituting the values we have

z = (0.62-0.60)/√{0.60(1-0.60)/1100}

z = 1.354

z = 1.35

To determine the p value (test statistic) at 0.01 significance level, using a two tailed hypothesis.

P value = P(Z<-1.35) + P(Z>1.35) = 0.0885 + 0.0885= 0.177

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = 1.35) falls with the region bounded by Z at 0.1 significance level. And also the two-tailed hypothesis P-value is 0.177 which is greater than 0.1. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 0.10 significance level the null hypothesis is valid.

3 0
3 years ago
Factor 16x+40y I don't know how to do this
Marizza181 [45]

Answer:

8(2x + 5y)

Step-by-step explanation:

You cant fact (take out) 8 from both 16 and 40 since 8 is a factor for both.

4 0
3 years ago
Read 2 more answers
Suppose the value R(d) of d dollars in euros is given by R(d)-d The cost P(n) in dollars to purchase and ship n purses is given
S_A_V [24]

<u>Corrected Question</u>

Suppose the value R(d) of d dollars in euros is given by R(d)=\dfrac67 d. The cost in dollars to purchase and ship n purses is given by P(n)=66n+23. Write a formula for the cost, Q(n) in euros to purchase and ship n purses.

Answer:

Q(n)=\dfrac67(66n+23)

Step-by-step explanation:

The value R(d) of d dollars in euros is given by R(d)=\dfrac67 d

Therefore:

R(1)=\dfrac67

T$hat is, 1  dollars =\dfrac67$ euros

The cost P(n) in dollars to purchase and ship n purses is given by:

P(n) = 66n+23.

Therefore, the cost, Q(n) in euros to purchase and ship n purses

=\dfrac67 \cdot P(n)\\Q(n)=\dfrac67(66n+23)

8 0
3 years ago
What is the average of the following numbers: 18, 64, 23, 78, 82, 91
Katen [24]
To Find The Answer, We Need To Add All Of Them Together, Then Divide By the Number Of Values. So:
64+23+78+82+91
-----------------------
             5

<span>64+23+78+82+91 = 338.
</span>
338
-----
  5

338/5 = 67.6

So, The Average Is 67.6
3 0
3 years ago
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