State the Pythagorean Theorem in your own words

Answer:

(a) Null Hypothesis, [tex]H_0[/tex] : p = 71%   

    Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71%  

(b) The test statistics is 3.25.

(c) The p-value is 0.0006.

Step-by-step explanation:

We are given that a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed since then.

A survey of 110 college students reported that 91 of them work.

Let p = proportion of college students who work

(a) Null Hypothesis, [tex]H_0[/tex] : p = 71%   {means that % of college students who work is same as 71% since 2011}

Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 71%   {means that % of college students who work is different from 71% since 2011}

The test statistics that will be used here is One-sample z proportion statistics;

                                 T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of college students who reported they work = [tex]\frac{91}{110}[/tex] = 82.73%

           n = sample of students = 110

(b) So, test statistics  =  [tex]\frac{\frac{91}{110}-0.71}{\sqrt{\frac{\frac{91}{110}(1-\frac{91}{110})}{110} } }[/tex]

                                    =  3.25

The test statistics is 3.25.

(c) P-value of the test statistics is given by the following formula;

       P-value = P(Z > 3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                            = 1 - 0.99942 = 0.0006

So, the p-value is 0.0006.

The null and alternative hypotheses is [tex]\rm H_0:[/tex] p = 71% and [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%, the value test statistics is 3.25, and the p-value is 0.0006.and this can be determined by using the given data.

Given :

a) The hypothesis is given by:

Null hypothesis --   [tex]\rm H_0:[/tex] p = 71%

Alternate Hypothesis --  [tex]\rm H_a[/tex] : p [tex]\neq[/tex] 71%

b) The statistics test is given by:

[tex]\rm TS = \dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} }[/tex]

[tex]\rm TS = \dfrac{\dfrac{91}{110}-0.71}{\sqrt{\dfrac{\dfrac{91}{110}(1-\dfrac{91}{110})}{110}} }[/tex]

Simplify the above expression.

TS = 3.25

c) The p-value is given by:

P-value = P(Z>3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                = 1 - 0.99942

                                = 0.0006

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Answer:

C. P- value < 0.04 0.05

Step-by-step explanation:

hello,

we were given the sample size, n = 200

also the probability that the psychic correctly identifies the symbol on the 200 card is

[tex]p=\frac{50}{200}= 0.25[/tex]

using the large sample Z- statistic, we have

[tex]Z=\frac{p- 0.20}{\sqrt{0.2(1-0.2)/200} }[/tex]

   = [tex]\frac{0.25-0.20}{\sqrt{0.16/200}}[/tex]

    = 1.7678

thus the P - value for the hypothesis test is P(Z > 1.7678) = 0.039.

from the above, we conclude that the P- value < 0.04, 0.05

Answer:

The standard error of the mean is 0.3216

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 188

Sample mean =

[tex]\bar{x}= 10.97\text{ lb}[/tex]

Sample standard deviation =

[tex]s = 4.41\text{ lb}[/tex]

We have to estimate the standard error of the mean.

Formula for standard error:

[tex]S.E = \dfrac{s}{\sqrt{n}}[/tex]

Putting values, we get,

[tex]S.E =\dfrac{4.41}{\sqrt{188}} = 0.3216[/tex]

Thus, the standard error of the mean is 0.3216

Answer:

Option A is the right answer choice.

Step-by-step explanation:

10 plus a negative number is the same as subtracting it as a positive

(a) The class interval that contains the median is [tex]\( 1 < h \leq 2 \).[/tex] (b)The estimated mean number of hours is 1.8 hours.

To analyze the data and answer the questions about the time 30 children spent on their tablets one evening, let's break it down step by step.

(a) The median is the value that separates the higher half from the lower half of the data. For 30 children, the median position will be the average of the 15th and 16th values in the ordered data set.

Given the frequencies:

- [tex]\( 0 < h \leq 1 \): 6[/tex] children

- [tex]\( 1 < h \leq 2 \):[/tex] 13 children

- [tex]\( 2 < h \leq 3 \):[/tex] 7 children

- [tex]\( 3 < h \leq 4 \): 4[/tex] children

To find the median class interval, we need to determine where the 15th and 16th values fall.

Cumulative frequency:

[tex]- \( 0 < h \leq 1 \): 6\\ - \( 1 < h \leq 2 \): 6 + 13 = 19\\ - \( 2 < h \leq 3 \): 19 + 7 = 26\\ - \( 3 < h \leq 4 \): 26 + 4 = 30[/tex]

From the cumulative frequency:

The 15th and 16th values fall within the second class interval [tex]\( 1 < h \leq 2 \),[/tex] because the cumulative frequency reaches 19 in this interval.

The class interval that contains the median is [tex]\( 1 < h \leq 2 \).[/tex]

(b) To estimate the mean, we'll use the midpoint of each class interval and multiply by the frequency of that interval, then divide by the total number of children.

Determine the midpoints of each class interval:

[tex]- \( 0 < h \leq 1 \): midpoint = \( \frac{0 + 1}{2} = 0.5 \)\\ - \( 1 < h \leq 2 \): midpoint = \( \frac{1 + 2}{2} = 1.5 \)\\ - \( 2 < h \leq 3 \): midpoint = \( \frac{2 + 3}{2} = 2.5 \)\\ - \( 3 < h \leq 4 \): midpoint = \( \frac{3 + 4}{2} = 3.5 \)[/tex]

Multiply each midpoint by its respective frequency:

[tex]- \( 0.5 \times 6 = 3 \)\\ - \( 1.5 \times 13 = 19.5 \)\\ - \( 2.5 \times 7 = 17.5 \)\\ - \( 3.5 \times 4 = 14 \)\\[/tex]

Sum these products:

[tex]\[ 3 + 19.5 + 17.5 + 14 = 54 \][/tex]

Divide by the total number of children to find the mean:

[tex]\[ \text{Mean} = \frac{54}{30} = 1.8 \][/tex]

The estimated mean number of hours is 1.8 hours.

The complete question is:

The table shows information about the numbers of hours 30 children spent on their tablets one evening.

Number of hours (m)          Frequency

    D<h<1                                   6

    1<h<2                                   13

    2<h<3                                   7

    3<h<4                                   4

a) Find the class interval that contains the median.

b) Work out an estimate for the mean number of hours.

Answer: 3rd option

Step-by-step explanation:

Answer:the third option is the answer

Step-by-step explanation:I just took the assignment on edge

Answer: 2.18

Step-by-step explanation:

log(11)/log(3)=2.18

Answer:

2.18

Step-by-step explanation:

I just did it and got it right

Answer: C

Step-by-step explanation:

Answer:

It might be C

Step-by-step explanation:

I am so sorry if it is the wrong answer. I didn't have a test about this.... :(

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

[tex]H_0: \mu=1102\\\\H_a:\mu < 1102[/tex]

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

[tex]H_0: \mu=1102\\\\H_a:\mu < 1102[/tex]

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

[tex]t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55[/tex]

The degrees of freedom are df=599

[tex]df=n-1=600-1=599[/tex]

The P-value for this test statistic is:

[tex]P-value=P(t<-2.55)=0.0055[/tex]

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

The mean is 21.9, the median is 22, the mode is 5 and the range is 43.

Important information:

Mean of the data set is:

[tex]Mean=\dfrac{5+28+16+32+5+16+48+29+5+35}{10}[/tex]

[tex]Mean=\dfrac{219}{10}[/tex]

[tex]Mean=21.9[/tex]

Arrange the data set in asccending order.

5, 5, 5, 16, 16, 28, 29, 32, 35, 48

The number of observation is 10, which is an even number. So, the median is average of [tex]\dfrac{10}{2}=5th[/tex] term and [tex]\dfrac{10}{2}+1=6th[/tex].

[tex]Median=\dfrac{16+28}{2}[/tex]

[tex]Median=\dfrac{44}{2}[/tex]

[tex]Median=22[/tex]

Mode is the most frequent value.

In the given data set 5 has the highest frequency 3. So, the mode of the data is 5.

Range is the data set is:

[tex]Range=Maximum-Minimum[/tex]

[tex]Range=48-5[/tex]

[tex]Range=43[/tex]

Therefore, the mean is 21.9, the median is 22, the mode is 5 and the range is 43.

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The requried x represents the number of items produced in each run, we should produce approximately 127 items in each run to minimize the total costs of production and storage.

To find the number of items that should be produced in each run to minimize the total costs of production and storage, we can use the following steps:

Let x be the number of items produced in each run. The total cost of production and storage (C) can be expressed as:

C(x) = (640/150,000) + (7x) + (0.75 * 150,000 / x)

The first term (640/150,000) represents the cost of preparing a production run, the second term (7x) represents the cost of producing each item, and the third term (0.75 * 150,000 / x) represents the cost of storing each item.

To minimize the total cost, we need to find the value of x that minimizes the function C(x). We can do this by taking the derivative of C(x) with respect to x and setting it equal to zero:

C'(x) = 0

Now, let's find the derivative of C(x):

C'(x) = d/dx [(640/150,000) + (7x) + (0.75 * 150,000 / x)]

C'(x) = 0 + 7 - (0.75 * 150,000 / x²)

Now, set C'(x) equal to zero and solve for x:

7 - (0.75 * 150,000 / x²) = 0

0.75 * 150,000 / x² = 7

150,000 / x² = 7 / 0.75

150,000 / x² = 9.333...

Now, solve for x:

x² = 150,000 / 9.333...

x² ≈ 16,071.43

x ≈ sqrt(16,071.43)

x ≈ 126.77

Since x represents the number of items produced in each run, we should produce approximately 127 items in each run to minimize the total costs of production and storage.

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Answer:

True

Step-by-step explanation:

Use the unit circle

tan(x) = 0 only when x is 0 or a multiple of 180 (in degrees)

Answer:

true

Step-by-step explanation:

Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

Check the conditions as follows:

 [tex]np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So,  [tex]\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)[/tex].

(1)

Compute the value of [tex]P(0.64<\hat p<0.69)[/tex] as follows:

[tex]P(0.64<\hat p<0.69)=P(\frac{0.64-0.65}{\sqrt{0.002275}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{0.69-0.65}{\sqrt{0.002275}})[/tex]

                              [tex]=P(-0.20<Z<0.80)\\=P(Z<0.80)-P(Z<-0.20)\\=0.78814-0.42074\\=0.3674[/tex]

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 90%.

That is,

[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]

Then,

[tex]P(p_{1}<\hat p<p_{2})=0.90[/tex]

[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.90[/tex]

[tex]P(-z<Z<z)=0.90\\P(Z<z)-[1-P(Z<z)]=0.90\\2P(Z<z)-1=0.90\\2P(Z<z)=1.90\\P(Z<z)=0.95[/tex]

The value of z for P (Z < z) = 0.95 is

z = 1.65.

Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex]  as follows:

[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57[/tex]                 [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73[/tex]

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let [tex]p_{1}[/tex] and [tex]p_{2}[/tex] be the two population percentages that will contain the sample percentage with probability 95%.

That is,

[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]

Then,

[tex]P(p_{1}<\hat p<p_{2})=0.95[/tex]

[tex]P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}<\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}})=0.95[/tex]

[tex]P(-z<Z<z)=0.95\\P(Z<z)-[1-P(Z<z)]=0.95\\2P(Z<z)-1=0.95\\2P(Z<z)=1.95\\P(Z<z)=0.975[/tex]

The value of z for P (Z < z) = 0.975 is

z = 1.96.

Compute the value of [tex]p_{1}[/tex] and [tex]p_{2}[/tex]  as follows:

[tex]-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55[/tex]                 [tex]z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75[/tex]

Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

The questions concern calculating the probability of a sample proportion falling within a given range and constructing confidence intervals around the population proportion. These questions are related to statistical concepts such as the normal approximation to the binomial distribution and the use of z-scores for interval estimation, which are typically covered in college-level statistics courses.

The student has asked about determining the probability that a sample percentage will fall within certain ranges, given a known proportion from a survey conducted by the Institute of Management Accountants (IMA) regarding the impact of global warming on their companies. Specifically, the student is looking to estimate probabilities related to the sample proportion and construct confidence intervals around a population percentage. These are statistical concepts typically covered in college-level courses in probability and statistics, specifically in chapters related to sampling distributions and confidence interval estimation.

To answer the first question, we would need to use the normal approximation to the binomial distribution, since the sample size is large (n=100). The sample proportion p = 0.65, and we can calculate the standard error for the sampling distribution of the sample proportion. However, since the full calculations are not provided here, a specific numerical answer cannot be given.

For the second and third questions, constructing confidence intervals at 90% and 95% requires using the standard error and the appropriate z-scores that correspond to these confidence levels. Again, the specific limits are not calculated here, but the process involves multiplying the standard error by the z-score and adding and subtracting this product from the sample percentage.

Step-by-step explanation:

Assume that

[tex]X_i = \left \{ {{1, If , Ith, room, has,exactly, 1,man, and , 1,woman } \atop {0, othewise} \right.[/tex]

hence,

[tex]x = x_1 + x_2+....+x_m[/tex]

now,

[tex]E(x) = E(x_1+x_2+---+x_m)\\\\E(x)=E(x_1)+E(x_2)+---+E(x_m)[/tex]

attached below is the complete solution

Answer:

We can see the details in the pic.

Step-by-step explanation:

We can see the details in the pic shown.

Answer:

[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]    

[tex]p_v =2*P(z<-2.24)=0.0251[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=13275[/tex] represent the sample mean

[tex]\sigma=2345[/tex] represent the sample standard deviation

[tex]n=500[/tex] sample size  

[tex]\mu_o =68[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] 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 true mean is 13960, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 13690[/tex]  

Alternative hypothesis:[tex]\mu \neq 13690[/tex]  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex]  (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) the info given like this:  

[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]    

P-value

Since is a two sided test the p value would be:  

[tex]p_v =2*P(z<-2.24)=0.0251[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.

Answer: The additive inverse of -11/-13 is 11/13

Step-by-step explanation: The additive inverse is the opposite sign. (positive or negative). It's the amount that needs to be added or subtracted to get 0.

Answer:

increases

Step-by-step explanation:

Answer:

the answer is:B

Step-by-step explanation:

Answer:

1.  [tex]l \leq 45[/tex] ( in this inequality, the time can be less than or equal to 45, but no more than 45)

2. [tex]m > 60[/tex]  (Mario's height is more than 60.)

3. [tex]f > 8000[/tex] (More than 8000 fans attended.)

Answer:

[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]

Step-by-step explanation:

Given

The jogger’s position: p(t)

We can express the acceleration a as follows

[tex]a=\frac{d^{2}p}{dt^{2}}[/tex]

then

[tex]\frac{d^{2}p}{dt^{2}}=k*\sqrt{t}[/tex]

only if

 [tex]0 min\leq t\leq 1 min[/tex]

The required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].

Given information:

A jogger runs along a straight track. The jogger’s position is given by the function p(t), where t is measured in minutes since the start of the run.

During the first minute of the run, the jogger’s acceleration is proportional to the square root of the time.

Let a be the acceleration of the jogger.

So, the expression for acceleration can be written as,

[tex]a=\dfrac{d}{dt}(\dfrac{dP(t)}{dt})\\a=\dfrac{d^2P(t)}{dt^2}[/tex]

Now, the acceleration is proportional to square root of time.

So,

[tex]a\propto \sqrt t\\a=k\sqrt t\\\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex]

Therefore, the required differential equation will be [tex]\dfrac{d^2P(t)}{dt^2}=k\sqrt t[/tex] or [tex]\dfrac{d^2P(t)}{dt^2}-k\sqrt t=0[/tex].

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Answer:

The correct option is (a).

Step-by-step explanation:

The complete question is:

Bankers at a large financial institution created the linear regression model dˆ=0.37−0.0004s to predict the proportion of customers who would default on their loans, d , based on the customer’s credit score, s . For a customer with a credit score of 700, which of the following is true?

(a) The default proportion is predicted to be 0.09.

(b) The default proportion will be 0.09.

(c) The default proportion is predicted to be approximately 1.75 million.

(d) The default proportion will be approximately 1.75 million.

(e) The default proportion is predicted to be 0.28.

Solution:

The linear regression model is used to predict the value of the response or dependent variable based on only one explanatory or independent variable.

The general form of a linear regression model is:

[tex]\hat y=\alpha +\beta x[/tex]

Here,

y = dependent variable

x = independent variable

α = intercept

β = slope

The linear regression model to predict the proportion of customers who would default on their loans, based on the customer’s credit score is:

[tex]\hat d=0.37-0.0004\ s[/tex]

d = default on loans

s = customer’s credit score

Compute the predicted value of d for s = 700 as follows:

[tex]\hat d=0.37-0.0004\ s[/tex]

  [tex]=0.37-(0.0004\times 700)\\=0.37-0.28\\=0.09[/tex]

Thus, for a customer with a credit score of 700, the default proportion is predicted to be 0.09.

Thus, the correct option is (a).

Answer:

on the top the negative rational numbers are placed

Answer:

On a horizontal number line, the positive rational numbers are placed To the right of, To the left of, above or below zero, and the negative rational numbers are placed To the right of, To the left of, above or below zero.

Answer: I am pretty sure it is 7

Step-by-step explanation:

the smaller one is to smaller than the bigger one so you would just add two to the 5

Answer:

7.5

Step-by-step explanation:

The triangles are proportional, so 6/4=x/5

1.5=x/5

x=1.5*5=7.5

Answer:

We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to​ 10%.

Step-by-step explanation:

We want to use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to​ 10%.

We set up our hypothesis to get:

[tex]H_0:p=0.10[/tex]------->null hypothesis

[tex]H_1:p\ne0.10[/tex]------>alternate hypothesis

This means that: [tex]p_0=0.10[/tex]

Also, we have that, one restaurant had 36 orders that were not accurate among 324 orders observed.

This implies that: [tex]\hat p=\frac{36}{324}=0.11[/tex]

The test statistics is given by:

[tex]z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n} } }[/tex]

We substitute to obtain:

[tex]z=\frac{0.11-0.1}{\sqrt{\frac{0.1(1-0.1)}{324} } }[/tex]

This simplifies to:

[tex]z=0.6[/tex]

We need to calculate our p-value.

P(z>0.6)=0.2743

Since this is a two tailed test, we multiply the probability by:

The p-value is 2(0.2723)=0.5486

Since the significance level is less than the p-value, we fail to reject the null hypothesis.

We do not have sufficient evidence to reject the claim that ,the rate of inaccurate orders is equal to​ 10%.

Answer:

My answer is not correct, do not read

Answer:

D (1,1)

Step-by-step explanation:

Answer:

D. (1,1)

Hope it works!

Step-by-step explanation:

Answer:

0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 8

We are given that the distribution of number of rushing yards per game is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(running back has 64 or fewer rushing yards)

[tex]P( x \leq 64) = P( z \leq \displaystyle\frac{64 - 50}{8}) = P(z \leq 1.75)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x \leq 64) = 0.9599[/tex]

0.9599 is the probability that a randomly selected running back has 64 or fewer rushing yards.

Answer:

x = 1

Step-by-step explanation:

We need to solve for x by isolating the variable.

First, expand the parentheses:

2(x + 1) = 4

2 * x + 2 * 1 = 4

2x + 2 = 4

Then subtract by 2:

2x + 2 - 2 = 4 - 2

2x = 2

Finally, divide by 2:

2x/2 = 2/2

x = 1

Thus, x = 1.

Hope this helps!

Answer:

x = 1

Step-by-step explanation:

Wellllll....

X = the number to solve the equation to equal 4

2(x +1) = 4

Multiply 2 by x and 1 by x

2x +2 = 4

Solve!

2x = 4-2

2x = 2

x = 1

Hope this works!

Rosa

Answer:

The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

All songs

Mean 215 seconds, standard deviation 35 seconds

Sample

49

Mean 215, standard deviation [tex]s = \frac{35}{\sqrt{49}} = 5[/tex]

The mean of the sampling distribution is of 215 seconds and the standard deviation is 5.

Answer:

The sample mean would be:

[tex]\mu_{\bar X} = 215 seconds[/tex]

And the deviation:

[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

We know the following info for the random variable X who represent the duration

[tex]\mu = 215, \sigma=35[/tex]

For this case we select a sampel size of n =49>30. So we can apply the central limit theorem. From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The sample mean would be:

[tex]\mu_{\bar X} = 215 seconds[/tex]

And the deviation:

[tex]\sigma_{\bar X} = \frac{35}{\sqrt{49}}= 5 seconds[/tex]