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Mashcka [7]
3 years ago
15

Enter the correct answer in the box

Mathematics
1 answer:
hjlf3 years ago
6 0
Divide 128 by 6 and that should help you with the problem then mutinying
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Solve y = x + 8 for x.<br> x= y + 8<br> x=y - 8<br> x= -y + 8<br> x = -y - 8
Anna007 [38]
X = y - 8

Step by step
Y = x + 8
x + 8 = y
-8 -8
x = y - 8

3 0
3 years ago
The manager of the quality department for a tire manufacturing company wants to study the average tensile strength of rubber use
Alika [10]

Answer:

p_v =P(z>z_{calc})=0.067  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis at 10% of significance.

So then the answer would be

a. true

Step-by-step explanation:

Data given and notation  

\bar X represent the sample mean

\sigma represent the population standard deviation

n sample size  

\mu_o =800 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less or equal than 800 for the null hypothesis:  

Null hypothesis:\mu \leq 800  

Alternative hypothesis:\mu > 800  

The z statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) and we assume that she got a calculated value z_{calc}

P-value

Since is a right tailed test the p value would be:  

p_v =P(z>z_{calc})=0.067  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis at 10% of significance.

So then the answer would be

a. true

3 0
4 years ago
Find the value of:<br><img src="https://tex.z-dn.net/?f=%20%5Crm%20%5C%3A%20%20%5Cfrac%7B2%7D%7B7%7D%20%20%5C%3A%20of%20%5C%3A%2
const2013 [10]

Answer:

2 days

Step-by-step explanation:

7 days =1 week

therefore

2/7=2 days

3 0
3 years ago
Read 2 more answers
A ball is launched into the sky at 54.4 feet per second from a 268.8 meter tall building. The equation for the ball’s height, h,
worty [1.4K]

Answer:

7.41

Step-by-step explanation:

Vertical component:

S = 268.8

U = 0

V =?

A = 9.8

T=?

S = ut + 0.5at^2. u = 0

268.8 = 0.5x9. 8xt^2

T = 7.41 seconds

5 0
3 years ago
Read 2 more answers
Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

5 0
4 years ago
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