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IrinaK [193]
3 years ago
12

Math (10 points) help please

Mathematics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

I would help but I am not good with graphing.

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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
4 years ago
1000000000000000000000000000000000000000000000X100/1/222X1
jok3333 [9.3K]

Answer:

4.5045045 x 10^44

Step-by-step explanation:

I just put it into a calculator idk if its even right

5 0
3 years ago
Find the missing side of the right triangle
vlada-n [284]

Answer:

a {}^{2}  + b {}^{2}  = c {}^{2}  \\ b {}^{2} =  \sqrt{c {}^{2}   -  a {}^{2} }  \\ b =  \sqrt{14 {}^{2}   - 3 {}^{2}  }  \\ b =  \sqrt{196 -  9}  \\ b =  \sqrt{187}  \\ b = 13.7

I hope this help

6 0
3 years ago
A freight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hours. What’s th
Agata [3.3K]
The original speed of the train was 25mph.
7 0
4 years ago
En una hoja de papel cuyo perímetro es de 96 centímetros, se quiere imprimir un volante de manera que el área impresa sea un rec
elixir [45]

Answer:

El perímetro de la región impresa es 72 cm y su área es 288 cm².  

Step-by-step explanation:

1. Tenemos el perímetro de la hoja de papel:

P₁ = 96 cm = 2l₁ + 2a₁  (1)  

Como sabemos el margen superior, inferior, izquierdo y derecho podemos encontrar la relación entre el largo y ancho del rectángulo interno (región impresa) con el largo (l) y ancho (a) del rectángulo externo (hoja de papel):      

l_{2} = l_{1} - (m_{s} + m_{i}) = l_{1} - (3 cm + 2 cm) = l_{1} - 5 cm  (2)            

a_{2} = a_{1} - (m_{d} + m_{iz}) = a_{1} - (2 cm + 5 cm) = a_{1} - 7 cm   (3)    

El perímetro del rectángulo interno es:

P_{2} = 2l_{2} + 2a_{2}    (4)

Introduciendo la ecuación (2) y (3) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2(l_{1} - 5 cm) + 2(a_{1} - 7 cm) = 2l_{1} + 2a_{1} - 10 cm - 14 cm = 96 cm - 24 cm = 72 cm  

Por lo tanto el perímetro del rectángulo interno (región impresa) es 72 cm.

 

2. Ahora para encontrar el área rectángulo interno debemos encontrar el largo y ancho del mismo, sabiendo que:

l_{2} = 2a_{2}     (5)

Introduciendo (5) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2*2a_{2} + 2a_{2} = 6a_{2}

a_{2} = \frac{P_{2}}{6} = \frac{72 cm}{6} = 12 cm

l_{2} = 2a_{2} = 2*12 cm = 24 cm

Entonces el área es:

A_{2} = l_{2}*a_{2} = 12 cm*24 cm = 288 cm^{2}

Por lo tanto el área del rectágulo interno (región impresa) es 288 cm².      

Espero que te sea de utilidad!  

7 0
3 years ago
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