Evaluate: 955-105 x 4 +118

Answer:

3 inches you are right

Step-by-step explanation:

Answer: E=1

Step-by-step explanation:

because 2 times 1 is 2 or any number multiplied by 1 is itself

Answer:

e=1, e= -1

Step-by-step explanation:

Case 1:

2e=2

e=2/2

e=1

Case 2:

2e= -2

e= -2/2

e= -1

Hope this helped!!!

4 divided by 1/2 as a fraction is equal to 8/1 or simply 8.

To divide a number by a fraction, you can multiply the number by the reciprocal of the fraction.

In this case, we have 4 divided by 1/2. The reciprocal of 1/2 is 2/1 (or simply 2).

So, we can rewrite the expression as:

4 * (2/1)

Multiplying these numbers, we get:

8/1

Therefore, 4 divided by 1/2 as a fraction is equal to 8/1 or simply 8.

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Final answer:

The circumference of a circle with an 80-inch radius is 502.4 inches, calculated using the formula C = 2πr with π approximated as 3.14.

Explanation:

The circumference of a circle with a radius of 80 inches can be calculated using the formula C = 2πr, where π (Pi) is approximately 3.14 and r is the radius. For given radius of 80 inches, the calculation would be:

C = 2 × 3.14 × 80 inches

C = 502.4 inches.

Therefore, the circumference of the circle is 502.4 inches.

Final answer:

The probability that a randomly chosen student from the Algebra 2 class plays basketball or baseball is 13/22, which is approximately 0.59 or 59% when calculated using the principle of inclusion and exclusion in probability.

Explanation:

To calculate the probability that a student chosen randomly from the Algebra 2 class plays basketball or baseball, you need to use the principle of inclusion and exclusion in probability theory. The total number of students in the class is 22. There are 5 students who play basketball and 11 who play baseball, but since 3 students play both sports, these students have been counted twice in the total of 5+11=16. Therefore, the correct number of students who play at least one of the sports is 5+11-3=13.

The probability that a randomly chosen student plays either basketball or baseball is hence the number of students who play either one or both sports divided by the total number of students in the class. This gives us 13/22. To convert this fraction to a decimal, we can divide 13 by 22, which gives us approximately 0.59. Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is approximately 0.59, or 59%.

Answer:

[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of​ girls

The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.

Step-by-step explanation:

A number x is said to be significantly high if:

[tex]P(X \geq x) \leq 0.05[/tex]

The appropriate probability to determine if a value x is significantly high is the probability of being equal or higher than x. It this probability is 0.05 or less, x is significantly high.

Is 55 girls in 100 births a significantly high number of​ girls?

We have that:

[tex]P(X \geq 55) = 0.184 \geq 0.05[/tex], so 55 girls in 100 births is not a significantly high number of​ girls

Answer:

a) [tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel and we got:

[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]

b) [tex]z=-1.036<\frac{a-122}{18}[/tex]

And if we solve for a we got

[tex]a=122 -1.036*18=103.35[/tex]

So the value of height that separates the bottom 15% of data from the top 85% is 103.35.  

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(122,18)[/tex]  

Where [tex]\mu=122[/tex] and [tex]\sigma=18[/tex]

We are interested on this probability

[tex]P(X>140)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel and we got:

[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]

Part b

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X<a)=0.15[/tex]   (a)

[tex]P(X>a)=0.85[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.15 of the area on the left and 0.85 of the area on the right it's z=-1.036. On this case P(Z<-1.036)=0.15 and P(z>-1.036)=0.85

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.15[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.15[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.036<\frac{a-122}{18}[/tex]

And if we solve for a we got

[tex]a=122 -1.036*18=103.35[/tex]

So the value of height that separates the bottom 15% of data from the top 85% is 103.35.  

Answer:

[tex]T(t)=-50e^{-0.4t}+42[/tex]

Step-by-step explanation:

We are given that

[tex]T(0)=-8^{\circ} C[/tex]

[tex]T'(t)=20e^{-0.4t}[/tex]

Integrating on both sides

[tex]T(t)=20\int e^{-0.4 t}[/tex]

[tex]T(t)=\frac{20}{-0.4}e^{-0.4t}+C[/tex]

Using the formula

[tex]\int e^{ax}dx=\frac{e^{ax}}{a}+C[/tex]

Substitute t=0 and  T(0)=-8

[tex]-8=-50+C[/tex]

[tex]C=-8+50=42[/tex]

Substitute the value of C

[tex]T(t)=-50e^{-0.4t}+42[/tex]

Answer:

D(1)=300

D(n)=d(n-1)× 1/5

Step-by-step explanation:

The 1st term is 300 and the common ratio is 1/5. You have to multiply times 1/5 to get to the next number.

Answer:

300

1/5

Step-by-step explanation:

Answer:

E(X) = 1.5

Var(X) = 2.325

Step-by-step explanation:

X - the number of pairs of cats (out of the 5 cats) to choose the same type of treat, can take values of:

X = (0, 1, 2, 3, 4, 5)

Also probability of choosing a treat, since they are all equally likely is: f(x) = 1/10

E(X) - expectation of x, is given by:

E(X) = Summation [X*f(x)]

E(X) = 0x(1/10) + 1x(1/10) + 2x(1/10) + 3x(1/10) + 4x(1/10) + 5x(1/10)

= (1/10) x (1+2+3+4+5)

E(X) = 3/2 = 1.5

Also, variance is:

Var(X) = Summation f(x)*[X - E(X)]^2

= (1/10)x(0-1.5)^2 + (1/10)x(1-1.5)^2 +(1/10)x(2-1.5)^2 +(1/10)x(3-1.5)^2 +(1/10)x(4-1.5)^2 +(1/10)x(5-1.5)^2

= (1/10)x[2.25 + 0.25 + 2.25 + 6.25 + 12.25]

Var(X) = 2.325

The value of the sample mean is 12.

Given

A sample with a mean of n = 9 has [tex]\rm \sum x=108[/tex].

The formula is used to calculate sample mean is;

[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\[/tex]

Substitute all the values in the formula

Then,

The sample mean is;

[tex]\rm Sample \ mean = \dfrac{\sum x}{n}\\\\\rm Sample \ mean = \dfrac{108}{9}\\\\ \rm Sample \ mean = 12[/tex]

Hence, the value of the sample mean is 12.

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Answer:

a) The probability that exactly three of them fail the test is 2.61%.

b) The probability that fewer than three of them fail the test is 97.02%.

c) The probability that more than two of them fail the test is 2.98%.

d) It would not be unusual for none of them to fail the test

Step-by-step explanation:

For each automobile, there are only two possible outcomes. Either it fails the test, or it does not. The probability of an automobile faiiling the test is independent of other automobiles. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

8% fail the emissions test.

This means that [tex]p = 0.08[/tex]

Nine automobiles.

This means that [tex]n = 9[/tex]

a. Find the probability that exactly three of them fail the test.

This is [tex]P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{9,3}.(0.08)^{3}.(0.92)^{6} = 0.0261[/tex]

The probability that exactly three of them fail the test is 2.61%.

b. Find the probability that fewer than three of them fail the test.

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{9,0}.(0.08)^{0}.(0.92)^{9} = 0.4722[/tex]

[tex]P(X = 1) = C_{9,1}.(0.08)^{1}.(0.92)^{8} = 0.3695[/tex]

[tex]P(X = 2) = C_{9,2}.(0.08)^{2}.(0.92)^{7} = 0.1285[/tex]

[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.4722 + 0.3695 + 0.1285 = 0.9702[/tex]

The probability that fewer than three of them fail the test is 97.02%.

c. Find the probability that more than two of them fail the test.

Either fewer than three(two or less) fail, or more than two do. The sum of the probabilities of these events is 100%. So

97.02 + p = 100

p = 2.98

The probability that more than two of them fail the test is 2.98%.

d. Would it be unusual for none of them to fail the test?

More than 2.5 standard deviations from the mean is unusual.

[tex]E(X) = np = 9*0.08 = 0.72[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{9*0.08*0.72} = 0.81[/tex]

0 > 0.72 - 2*0.81

So

It would not be unusual for none of them to fail the test

To find the probabilities, we use the binomial probability formula. For each part of the question, we plug in the appropriate values and calculate the probabilities. To determine if it would be unusual for none of them to fail, we calculate the probability of exactly zero failures.

To find the probabilities in this question, we can use the binomial probability formula: P(x) = (nCx) * p^x * (1-p)^(n-x), where n is the number of trials (automobiles selected), x is the number of successes (fail the test), and p is the probability of success (8% or 0.08).

a. For exactly three failures, we use x = 3 and n = 9 in the formula. P(3) = (9C3) * 0.08^3 * (1-0.08)^(9-3).

b. For fewer than three failures, we find the probabilities of 0, 1, and 2 failures and sum them up: P(<3) = P(0) + P(1) + P(2).

c. For more than two failures, we find the probabilities of 3, 4, 5, 6, 7, 8, and 9 failures and sum them up: P(>2) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9).

d. To determine if it would be unusual for none of them to fail, we calculate P(0) using x = 0 and n = 9 in the formula. If P(0) is very low, it would be considered unusual.

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It's the same as any other shape. Find the lengths of the sides and add them up.

Answer:

To find the perimeter of a quadrilateral you would add the lengths of each  side together.

Example:

The perimeter is the sum or all the lengths.

If there was a rectangle with a length of 5 cm and a width of 3 cm.

To work this out you would first multiply 5 by 2, which is 10. Then you would multiplying 3 by 2, which is 6. Then you would add 6 and 10, which is 16.

1) Multiply 5 By  2.

[tex]5*2=10[/tex]

2) Multiply 3 by 2.

[tex]3*2=6[/tex]

3) Add 10 and 6.

[tex]10+6=16[/tex]

9514 1404 393

Answer:

 (a) Pyramid 1 by 58.6 square units

Step-by-step explanation:

The area of each triangle is given by the formula ...

  A = 1/2bh

The total area is the area of all four triangles that make up the outer surface of the pyramid.

Pyramid 1

  A = 1/2(12×2.5 +12×13.1 +2(6.5×14)) = 184.6 . . . square units

Pyramid 2

  A = 1/2(12×4.5 +12×6.5 +2(7.5×8)) = 126 . . . square units

Pyramid 1 has the larger area by ...

  184.6 -126 = 58.6 . . . square units

The ratio 3:x equivalent to 1:9 results in x being equal to 27.

To find the number that makes the given ratio equivalent to 1:9 we can set up a proportion using the information given which is a ratio of 3 to an unknown number (let's call it x).

First we write down the two ratios as fractions and set them equal to each other to find x:

1/9 = 3/x

We then cross-multiply to solve for x:

1 × x = 9 × 3

This gives us:

x = 27

So, the number that makes the ratio equivalent to 1:9 when compared to 3 is 27.

Answer:

23400 cm

Step-by-step explanation:

30 cm x 30 cm x 26 cm = 23,400 cm

Answer:

$225

Step-by-step explanation:

-RevPar is defined as the room revenue divided by the number of rooms available:

[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}[/tex]

-The average number of rooms sold per day in the two days is:

[tex]mean =\frac{450}{2}\\\\=225\ rooms[/tex]

The PevPar can then be calculated as:

[tex]RevPar=\frac{Room \ Revenue}{Rooms \ Available}\\\\=\frac{ADR\times mean occupancy}{Rooms \ Available}\\\\=\frac{300\times 225}{300}\\\\=225[/tex]

Hence, the PevPar is $225

If the two sides with length 3 and 4 are the two legs, then the missing side, i.e. the hypotenuse, is indeed 5.

But it could also be the case that 4 is the hypotenuse and 3 is one of the legs. In this case, the missing side is the other leg, so we calculate it using

[tex]\sqrt{4^2-3^2}=\sqrt{16-9}=\sqrt{7}[/tex]

So, a right triangle with legs [tex]\sqrt{7}[/tex] and 3 has an hypotenuse of 4.

Moira's statement correctly applies the Pythagorean theorem, implying it is impossible to have a right triangle with sides of 3, 4, and a value other than 5 and still conform to the requirements of a right-angled triangle.

The Pythagorean theorem, where a2 + b2 = c2 applies, and c represents the length of the hypotenuse, while a and b represent the lengths of the other two sides. However, the question asked to provide an example that opposes Moira's claim by asking for a right triangle with sides of length 3, 4, but with a hypotenuse different from 5. This is a trick question because, according to the Pythagorean theorem, if a triangle has sides of 3 and 4 units, the hypotenuse must indeed be 5 units to satisfy the equation 32 + 42 = 52 (9 + 16 = 25). Therefore, there cannot be a right-angle triangle with the sides 3, 4, and a number other than 5, as it would violate the Pythagorean theorem.

Answer:

6x∧4 + 14x³ - 12x²

Step-by-step explanation:

2x²(3x² + 7x - 6)

6x∧4 + 14x³ - 12x²

Tell me if I am wrong.

Can I get brainliest

Answer:the answer is B

Step-by-step explanation:

Answer:

B  0 (0.3] U (9,18); only (9,18] contains viable times for machine 1

Step-by-step explanation:

The graph is 0<x<3 or 9<x<18

Answer:

30i

Step-by-step explanation:

[tex]5 \sqrt{ - 4} + 4 \sqrt{ - 25} \\ = 5 \sqrt{4 \times - 1} + 4 \sqrt{25 \times - 1} \\ = 5 \sqrt{4} \times \sqrt{ - 1} + 4 \sqrt{25} \times \sqrt{ - 1} \\ = 5 \times 2 \times i + 4 \times 5 \times i...( \because \sqrt{ - 1} = i) \\ = 10i + 20i \\ = 30i[/tex]

Answer:

a) For this case we select a sample size of n =9. And the distribution for the sample mean is given by:

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

With the following parameters:

[tex]\mu_{\bar X}= 96[/tex]

[tex]\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767[/tex]

b) [tex] P(95< \bar X < 100)[/tex]

And we can use the z score formula given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And if we find the z score for the limit we got:

[tex] z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129[/tex]

[tex] z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515[/tex]

[tex] P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248[/tex]

The sketch is on the figure attached

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(96,23.3)[/tex]  

Where [tex]\mu=96[/tex] and [tex]\sigma=23.3[/tex]

Part a

For this case we select a sample size of n =9. And the distribution for the sample mean is given by:

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

With the following parameters:

[tex]\mu_{\bar X}= 96[/tex]

[tex]\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767[/tex]

Part b

For this case we want this probability:

[tex] P(95< \bar X < 100)[/tex]

And we can use the z score formula given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And if we find the z score for the limit we got:

[tex] z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129[/tex]

[tex] z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515[/tex]

[tex] P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248[/tex]

The sketch is on the figure attached

Answer:

[tex]x = 8\\y=-7[/tex]

Step-by-step explanation:

[tex]-3y-4x=-11\\3y-5x=-61[/tex]

In order to eliminate one of the variables you can add both equations.

[tex]-9x=-72\\x=\frac{-72}{-9} \\x=8[/tex]

Replace in one of the main equations to find y

[tex]3y-5x=-61\\3y-5(8)=-61\\3y-40=-61\\3y=-61+40\\3y=-21\\y=\frac{-21}{3} \\y=-7[/tex]

Replace in one of the main equations to prove that your answers are correct.

[tex]-3y-4x=-11\\-3(-7)-4(8)=-11\\21-32=-11\\-11=-11[/tex]

Answer:

Correct option is (d).

Step-by-step explanation:

A hypothesis test for single mean can be used to determine the significance of the claimed value of the population mean μ.

Now there are two test for mean:

A z-test is used when it is provided that the population is normally distributed, the sample is large and the population standard deviation value is provided.

A t-test is used when it is assumed that the population is normally distributed, the sample is large enough and the population standard deviation is not known.

So, in case there is no information about the population standard deviation (σ) but the sample standard deviation is either given or can be calculated from the provided data set, then the hypothesis test for single mean can be performed using the t-test.

Hence, the choice between a z-test and a t-test for a population mean depends primarily on whether the given standard deviation is from the population or the sample.

Thus, the correct option is (d).

Answer:

There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The claim is that the cars belonging to students are, on average, older than the cars belonging to faculty.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0[/tex]

being μ1: population mean age of students cars, μ2: population mean age of faculty cars.

The significance level is assumed to be 0.05.

The information about the students cars sample is:

Mean M1: 8.5 years.

Standard deviation s1: 6.2 years.

Sample size n1: 58 cars.

The information about the faculty cars sample is:

Mean M2: 5.1 years.

Standard deviation s2: 3.5 years.

Sample size n2: 41 cars.

The difference between means is:

[tex]M_d=M_1-M_2=8.5-5.1=3.4[/tex]

The standard error of the difference between means is:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}} =\sqrt{\dfrac{6.2^2}{58}+\dfrac{3.5^2}{58}}=\sqrt{ 0.6628 +0.2988 }=\sqrt{0.9615}\\\\\\s_{M_d}=0.9806[/tex]

Now we can calculate the t-statistic as:

[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{3.4-0}{0.9806}=3.4673[/tex]

The degrees of freedom are 93.033791, so the P-value for this right tail test is:

[tex]P-value=P(t>3.4673)=0.0004[/tex]

As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.

There is evidence to support the claim that the cars belonging to students are, on average, older than the cars belonging to faculty.

Step-by-step explanation:

Given a builder makes drainpipes that drop 1 cm over a horizontal distance of 30cm to prevent clogs. The ratio of the vertical drop to the horizontal distance covered is [tex]\frac{1}{30}[/tex] cm.

The angle of inclination,

tan(θ) = [tex]\frac{1}{30}[/tex]

θ = 1.9º

By trignometric ratio,

cos(θ) = [tex]\frac{horizontal distance}{length of the drainpipe}[/tex]

length of the drainpipe = [tex]\frac{700}{cos(1.91)}[/tex] = [tex]\frac{700}{0.999}[/tex]

length of the drainpipe = 700.7 cm

Answer:

The correct Answer for Khan Academy is 700.4

Step-by-step explanation:

Answer:

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram   {means that the machine is under filling the bags}

Step-by-step explanation:

We are given that a manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 430 gram setting.

It is believed that the machine is under filling the bags. A 21 bag sample had a mean of 429 grams with a variance of 289.

Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\geq[/tex] 430 gram   {means that the machine is not under filling the bags}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 430 gram   {means that the machine is under filling the bags}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean = 429 grams

             s = sample standard deviation = [tex]\sqrt{289}[/tex] = 17 grams

             n = sample of bags = 21

So, test statistics  =  [tex]\frac{429-430}{\frac{17}{\sqrt{21} } }[/tex]  ~ [tex]t_2_0[/tex]   

                               =  -0.269

Now at 0.01 significance level, the t table gives critical value of -2.528 at 20 degree of freedom for left-tailed test. Since our test statistics is higher than the critical value of t as -0.269 > -2.528, 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 machine is not under filling the bags.

Final answer:

The student's question involves setting up null and alternative hypotheses for a hypothesis test related to a bag filling machine's weight accuracy. The null hypothesis posits no difference from the 430-gram setting, while the alternative hypothesis suggests underfilling. The test uses a 0.01 significance level to determine whether to reject the null hypothesis.

Explanation:

The student is asking about hypothesis testing in the context of quality control for a bag filling machine. The null hypothesis (H0) assumes that there's no difference between the sample's mean weight and the population mean weight that the machine is set to, which is 430 grams; hence H0:
μ = 430 grams. The alternative hypothesis (Ha) suggests that the actual mean weight is less than the setting, given the concern about underfilling; thus Ha: μ < 430 grams.

For the given scenario, the bag sample's mean is 429 grams with a variance of 289 grams. Since a 0.01 level of significance will be used, if the test statistic calculated from the sample data falls into the critical region determined by this significance level, the null hypothesis would be rejected in favor of the alternative hypothesis.

Final answer:

The solution to the equation x^3 = 25 involves taking the cube root of both sides, yielding an approximate solution of x ≈ 2.924, rounded to three decimal places.

Explanation:

The question asks to find the solution to the equation x^3 = 25. This equation requires us to find a value of x such that, when cubed, equals 25. To solve this, we take the cube root of both sides of the equation, which gives us x = √[3]{25}. This cube root can be simplified using a calculator or estimated for a rough solution.

Thus, the solution to the equation is x ≈ 2.924 (rounded to three decimal places). It's important to note that in this case, we are only looking for real numbers as solutions. Complex solutions aren't considered for this specific problem.

The solution to the equation x^3 = 25 is found by taking the cube root of both sides. The approximate numerical solution is x ≈ 2.924.

The solution to the equation x^3 = 25 involves finding a number x that, when cubed, equals 25. We can solve this by taking the cube root of both sides of the equation. This operation will give us the principal root of x. The cube root of 25 is not a whole number, so we will end up with a decimal answer. To solve for x, we proceed as follows:

The actual value of x is a decimal number that, because of the nature of cube roots, cannot be expressed as an exact fraction or simple square root. The approximate numerical solution to the given cubic equation is x ≈ 2.92401773821287.

The complete question is;

A tank contains 60 lb of salt dissolved in 400 gallons of water. A brine solution is pumped into the tank at a rate of 4 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 4 gal/min. Determine the amount of salt in the tank at any time

t, if the concentration in the inflow is variable and given by

c(t) = 2 + sin(t/4) lb/gal.

Answer:

dA/dt = (8 + 4sin(t/4)) - (A_t/100)

Step-by-step explanation:

Rate is given as;

dA/dt = R_in - R_out

R_in = (concentration of salt inflow) x (input rate of brine)

So, R_in = (2 + sin(t/4)) x 4 = (8 + 4sin(t/4))

The solution is being pumped out at the same rate, thus it is accumulating at the same rate.

After t minutes, there will be 400 + (0 x t) gallons left = 400 gallons left

Thus,

R_out =(concentration of salt outflow) x (output rate of brine)

R_out = (A_t/400) x 4 = A_t/100

Thus,

A_t = (100/t)(8t - 16cos(t/4) - 60)

Since we want to find the amount of salt, A(t), let's integrate;

Thus, A = 8t - 16cos(t/4) - (A_t/100)t

A = 60 from the question. Thus,

60 = 8t - 16cos(t/4) - (A_t/100)t

Let's try to make A_t the subject;

(A_t/100)t = 8t - 16cos(t/4) - 60

A_t = (100/t)(8t - 16cos(t/4) - 60)

Answer:

Reject [tex]H_0[/tex] . The change is statistically significant. The software does appear to improve exam scores.

Step-by-step explanation:

We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.

In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.

Let [tex]\mu[/tex] = mean scores on the final exam.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex]  114 points   {means that the mean scores on the final exam does not increases after using software}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 114 points   {means that the mean scores on the final exam increase significantly after using software}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                       T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample average points = 117 points

            s = sample standard deviation = 8.7 points

            n = sample of students = 218

So, test statistics  =  [tex]\frac{117-114}{\frac{8.7}{\sqrt{218} } }[/tex]  ~ [tex]t_2_1_7[/tex]   

                               =  5.091

Now, P-value of the test statistics is given by the following formula;

         P-value = P( [tex]t_2_1_7[/tex] > 5.091) = Less than 0.05%

Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.

The test statistic is 3.448 and the P-value is 0.0014. The appropriate conclusion is that the software does appear to improve exam scores.

The test statistic can be calculated using the formula:

test_statistic = (sample_mean - population_mean) / (sample_standard_deviation / sqrt(sample_size))

Substituting in the given values:

test_statistic = (117 - 114) / (8.7 / sqrt(218)) = 3.448

The P-value can be calculated using a statistical software or calculator. For this question, the P-value is 0.0014

Since the P-value is less than 0.05 (significance level), we can reject the null hypothesis. The appropriate conclusion is: iv. Reject H_0. The change is statistically significant. The software does appear to improve exam scores.

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