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Lady_Fox [76]
2 years ago
12

Will give crown will post one more

Mathematics
2 answers:
zubka84 [21]2 years ago
6 0

Answer:

1 is 22; 2 is 19; 3 is 16; 4 is 13; 5 is 10

Step-by-step explanation:

Give me that crown again

Oksi-84 [34.3K]2 years ago
3 0

Answer:

22, 19, 16, 10

Step-by-step explanation:

y = 25-(3*1)= 22

y = 25-(3*2)= 19

y = 25-(3*3)= 16

y = 25-(3*5)= 10

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Select the graph that shows the correct sum
Lelechka [254]

Answer:B

Step-by-step explanation:

5 0
3 years ago
(m^(6)n^(3))/(m^(2)n^(6))
mario62 [17]

Answer:

(m^4)/(n^3)

Step-by-step explanation:

See attachment

7 0
2 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
Select the angle whose sine equals 12/37
Mazyrski [523]

Answer:

r

Step-by-step explanation:

sin is o/h so across from 12 would be the angle r

6 0
3 years ago
The difference of two numbers is 90 and their quotient is 10
bija089 [108]
To solve this, set up two equations using the information you're given. Let's call our two numbers a and b:
1) D<span>ifference of two numbers is 90
a - b (difference of two numbers) = 90

2) The quotient of these two numbers is 10
a/b (quotient of the two numbers) = 10


Now you can solve for the two numbers.
1) Solve the second equation for one of the variables. Let's solve for a:
a/b = 10
a = 10b

2) Plug a =10b into the first equation and solve for the value of b:
a - b = 90
10b - b = 90
9b = 90
b = 10

3) Using b = 10, plug it back into one of the equations to find the value of a. I'll plug it back into the first equation:
a - b = 90
a - 10 = 90
a = 100

-------

Answer: The numbers are 100 and 10</span>
3 0
3 years ago
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