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2 years ago

HELP BRAINLIEST FOR. CORRECT AND ALL ANSWERED ANSWER

Mathematics

2 answers:

UNO [17]2 years ago

8 0

Answer:

Step-by-step explanation:

SpyIntel [72]2 years ago

7 0

Answer:

1. 6/7

2. 1/2

3. 7/10

4. 39/40

5. 15/28

6. 3/10

7. 7/12

8. 1/6

9. 5/12

10. 1/8

Step-by-step explanation:

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Determine the x-intercept of the line whose equation is given:

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cause blue and red make orange :)

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George is 8 years older than Christopher and Ford is 2 years younger than Christopher. The sum of their ages is 60. Find the age

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Can I please get some help on this...I have tried to many times...

Bezzdna [24]

Answer: A

Step-by-step explanation:

First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.

$\sqrt{x-2} =0$

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3 years ago

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed

tatyana61 [14]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 8 minutes

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes

(b) The P-value of the test statistics is 0.0436.

(c) We conclude that the actual mean waiting time equals the standard at 0.05 significance level.

(d) 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Step-by-step explanation:

We are given that the length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes.

A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation of 3.2 minutes.

Let $\mu$ = actual mean waiting time.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 8 minutes {means that the actual mean waiting time does not differs from the standard}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 8 minutes {means that the actual mean waiting time differs from the standard}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

So, test statistics = $\frac{8.5-8}{\frac{3.2}{\sqrt{120} } }$

= 1.71

The value of t test statistics is 1.71.

(b) Now, the P-value of the test statistics is given by;

P-value = P(Z > 1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.9564 = 0.0436

(c) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual mean waiting time equals the standard.

(d) Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean waiting time = 8.5 minutes

$\sigma$ = population standard deviation = 3.2 minutes

n = sample of shoppers = 120

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population proportion, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $8.5-1.96 \times{\frac{3.2}{\sqrt{120} } }$ , $8.5+1.96 \times{\frac{3.2}{\sqrt{120} } }$ ]

= [7.93 minutes , 9.07 minutes]

Therefore, 95% confidence interval for the population mean is [7.93 minutes , 9.07 minutes].

Yes, it support our above conclusion.

3 0

3 years ago

Shelby Industries has a capacity to produce 45,000 oak shelves per year and is currently selling 40,000 shelves for $32 each. Ma

Dominik [7]

Answer:

y = mx + c

since m = 0

c = 9

$350,000 + $25.4 = y value

5 0

2 years ago