All categories

3 years ago

4. What percent of all computers sold during the entire week were sold on Friday!

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

Answer:

I confused... there are no numbers

You might be interested in

Một khối kim loại có thể tích 3,2cm3 và cân nặng 22,4g. Hỏi một khối kim loại cùng chất có thể tích là 4,5cm3 cân nặng bao nhiêu

Ymorist [56]

Answer:

Một cm3 của khối kim loại đó cân nặng số gam là:

22,4:3,2=7(g)

Một khối kim loại cùng chất có thể tích là 4,5cm3 cân nặng số gam là:

7x4,5=31,5(g)

Step-by-step explanation:

6 0

3 years ago

To test Upper H 0: muequals50 versus Upper H 1: muless than50, a random sample of size nequals23 is obtained from a populatio

natta225 [31]

Answer:

Step-by-step explanation:

Hello!

To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. 

 A) If ¯ x = 47.9 and s=11.9, compute the test statistic .

For thistest the corresponsing statistis is a one sample t-test

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}$

$t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85$

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

$t_{n-1;\alpha }= t_{22;0.1}= -1.321$

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If $t_{H_0}$ ≤ -1.321, reject the null hypothesis.

If $t_{H_0}$ > -1.321, do not reject the null hypothesis.

$t_{H_0}$ = -0.85, the decision is to not reject the null hypothesis.

To test H0: μ=100 versus H1:≠100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143$

b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

$t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807$

$t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807$

c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in thet-distribution?

Attachment.

(d) Will the researcher reject the null hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If $t_{H_0}$ ≤ -2.807 or if $t_{H_0}$ ≥ 2.807, reject the null hypothesis.

If -2.807 < $t_{H_0}$ < 2.807, do not reject the null hypothesis.

$t_{H_0}$ = 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80$

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

To resolve this you have to look for the values in the t-table that are the closest to the calculated $t_{H_0}$

Symbolically:

$t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}$

$t_{H_0}$ = -1.80

$t_{17; 0.025 }= -2.110$

$t_{17;0.05}= -1.740$

Roughly defined you can say that the p-value is the probability of obtaining the value of $t_{H_0}$ , symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0

3 years ago

Mark owns Siberian Husky sled dogs. He knows from data collected over the years that the weight of the dogs is a normal distribu

kozerog [31]

Calculate the z-score for the given data points in the item using the equation,

z-score = (x - μ) / σ

where x is the data point, μ is the mean, and σ is the standard deviation.

Substituting,
(47.7) z-score = (47.7 - 52.5)/2.4 = -2

This translates to a percentile of 2.28%.

(54.9) z-score = (54.9 - 52.5)/2.4 = 1

This translates to a percentile of 84.13%.

Then, subtract the calculate percentiles to give us the final answer of 81.85%.

Thus, 81.85% of the Siberian Husky sled dogs are expected to weigh between 47.7 and 54.9 lbs.

8 0

3 years ago

91. Jack can read 45 pages of his book in one and a

Digiron [165]

Answer:

10 hours

Step-by-step explanation:

45 pages per 1 hour and 30 minutes, or 60 min +30 min =90 min

90 minutes / 45 pages has to be equal to an equivalent fraction where we have 300 pages

90 minutes /45 pages = ? minutes / 300 pages , multiply both sides by 300

? minutes = 90*300/45 = 600 minutes

to read 300-page book will take 600 minutes = 10 hours

3 0

3 years ago

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-3,1) and (2,-1) is $y=-\frac{2}{5}x-\frac{1}{5}$

8 0

3 years ago