John has 1/3 as much money as grant. Grant gives John 9$. Now grant has 6$ more than John. How much money do grant and John have in all

Answer:

a) 0.3889

b) 0.5

c) 0.8333

d) The mean is 250 and the standard deviation is 51.96.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability of finding a value of X higher than x is:

[tex]P(X > x) = 1 - \frac{x - a}{b-a}[/tex]

The probability of finding a value of X between c and d is:

[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]

The mean and the standard deviation are, respectively:

[tex]M = \frac{a+b}{2}[/tex]

[tex]S = \sqrt{\frac{b-a}^{2}{12}}[/tex]

A random variable follows the continuous uniform distribution between 160 and 340.

This means that [tex]a = 160, b = 340[/tex]

a)

[tex]P(220 \leq X \leq 290) = \frac{290 - 220}{340 - 160} = 0.3889[/tex]

b)

[tex]P(160 \leq X \leq 250) = \frac{250 - 160}{340 - 160} = 0.5[/tex]

c)

[tex]P(X > 190) = 1 - \frac{190 - 160}{340 - 160} = 0.8333[/tex]

d)

[tex]M = \frac{160 + 340}{2} = 250[/tex]

[tex]S = \sqrt{\frac{340 - 160}^{2}{12}} = 51.96[/tex]

The mean is 250 and the standard deviation is 51.96.

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.

a. In words, define the random variable X

b. What is X ~N(,)

c. Find the probability that the child spends less than 2 hours per day unsupervised.

d. What percent of the children spend over 12 hours per day unsupervised?

Given Information:  

Mean = μ = 3.2 hours  

Standard deviation = σ = 1.7 hours  

Required Information:  

a. In words, define the random variable X

b. X ~N(,) = ?

c. P(X < 2) = ?

d. P(X > 12) = ?

Answer:  

a) X is the number of hours in a day that a 4-year-old child spends being unsupervised.

b) X ~N(μ,σ) = X ~N(3.2, 1.7)

c) P(X < 2) = 23.88%

d) P(X > 12) = 0%

Explanation:  

a)

Let X is the number of hours in a day that a 4-year-old child spends being unsupervised.

b)

X ~N(μ,σ) = X ~N(3.2, 1.7)

Where 3.2 is the average number of hours that 4-year-old child spends being unsupervised and 1.7 is the standard deviation.

c)

We want to find out the probability that a child spends less than 2 hours per day unsupervised.

P(X < 2) = P(Z < (x - μ)/σ)

P(X < 2) = P(Z < (2 - 3.2)/1.7)

P(X < 2) = P(Z < (- 1.2)/1.7)

P(X < 2) = P(Z < -0.71)

The z-score corresponding to -0.71 is 0.2388

P(X < 2) = 0.2388

P(X < 2) = 23.88%

Therefore, the probability that a child spends less than 2 hours per day unsupervised is 23.88%

d)

We want to find out the probability that a child spends over 12 hours per day unsupervised.

P(X > 12) = 1 - P(X < 12 )

P(X > 12) = 1 - P(X < (x - μ)/σ)

P(X > 12) = 1 - P(X < (12 - 3.2)/1.7)

P(X > 12) = 1 - P(X < 8.8/1.7)

P(X > 12) = 1 - P(X < 5.18)

The z-score corresponding to 5.18 is 1

P(X > 12) = 1 - 1

P(X > 12) = 0

Therefore, the probability that a child spends over 12 hours per day unsupervised is 0%

The question discusses unsupervised hours of Mercurian children and it's a statistics problem involving normal distribution where the average unsupervised time is 3.2 hours with a standard deviation of 1.7 hours.

This question is related to statistics, specifically about normal distribution. The mean unsupervised time for 4-year-olds on Mercury is 3.2 hours, with a standard deviation of 1.7 hours. If we randomly survey a 4-year-old Mercurian child living in a rural area, the time they likely spend alone, denoted as X, will range according to this distribution.

Normal distributions can be defined by two parameters: the mean (μ) and the standard deviation (σ). In this case, μ = 3.2 hours and σ = 1.7 hours. Therefore, we can say that the time X spent by this sampled child alone is normally distributed with these parameters.

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Given:

The height of the the tree from where the tree was broken = 13 ft

The distance from the foot of the tree to the broken top = 36 ft

To find the height of the tree.

Formula

By Pythagoras theorem we get,

h² = l²+b²,

where, h be the hypotenuse

b be the base and

l be the other side of the triangle along the right angle.

Now,

Putting, l =13 and b = 36 we get,

[tex]h^{2} =36^{2} +13^{2}[/tex]

or, [tex]h^{2} = 1296+169[/tex]

or, [tex]h^{2} =1465[/tex]

or, [tex]h=\sqrt{1465}[/tex]

or, [tex]h = 38.3[/tex]

Therefore,

The height of the tree is about = 13+38.3 ft = 51.3 ft

Hence,

The height of the tree is about 51.3 ft.

Answer:

k = 37

Step-by-step explanation:

[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]

answer:

1.94117647058

Step-by-step explanation:

33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058

Step-by-step explanation:

A health inspector must visit 3 of 13 restaurants on Monday.

In how many way can she pick 3 restaurants = 13 C 4 !

nCr ! = [tex]\frac{n!}{r!(n-r)!}[/tex]

13 C 4! = [tex]\frac{13!}{3!(10)!}[/tex]

= [tex]\frac{11.12.13}{2.3}[/tex]

= 286

Health inspector can pick 3 restaurants of 13 restaurants on Monday in 286 ways

Final answer:

There are 286 different ways for a health inspector to pick 3 out of 13 restaurants to visit, calculated using the combination formula C(n, k) = n! / (k!(n - k)!).

Explanation:

The student asked how many different ways a health inspector could pick 3 out of 13 restaurants to visit on Monday. This is a problem of combinatorics, specifically concerning combinations without repetition since the order of choosing restaurants does not matter.

To calculate this, we use the combination formula, which is defined as:

C(n, k) = n! / (k!(n - k)!)

Where:

Applying the formula:

C(13, 3) = 13! / (3!(13 - 3)!) = 13! / (3! * 10!) = (13*12*11) / (3*2*1) = 286

So, the health inspector has 286 different ways to pick 3 restaurants from a list of 13.

Given:

Given that the radius of the circle is 12 cm.

The length of the rectangle is 11 cm.

The width of the rectangle is 5 cm.

We need to determine the area of the shaded region.

Area of the rectangle:

The area of the rectangle can be determined using the formula,

[tex]A_1=length \times width[/tex]

Substituting the values, we have;

[tex]A_1=11\times 5[/tex]

[tex]A_1=55 \ cm^2[/tex]

Thus, the area of the rectangle is 55 square cm.

Area of the circle:

The area of the circle can be determined using the formula,

[tex]A_2=\pi r^2[/tex]

Substituting r = 12, we have;

[tex]A_2= (3.14)(12)^2[/tex]

[tex]A_2=452.16 \ cm^2[/tex]

Thus, the area of the circle is 452.16 square cm.

Area of the shaded region:

The area of the shaded region can be determined by subtracting the area of the rectangle from the area of the circle.

Thus, we have;

Area = Area of the circle - Area of the rectangle.

Substituting the values, we have;

[tex]Area=452.16-55[/tex]

[tex]Area=397.16 \ cm^2[/tex]

Thus, the area of the shaded region is 397.16 square cm.

By definition, if [tex]\vec v[/tex] is an eigenvector of [tex]\mathbf A[/tex] (with associated eigenvalue 8), then

[tex]\mathbf A\vec v=8\vec v[/tex]

[tex]\mathbf A^4\vec v=\mathbf A^3\cdot\mathbf A\vec v=8\mathbf A^3\vec v[/tex]

Rinse and repeat to find

[tex]\mathbf A^4\vec v=8^4\vec v[/tex]

so that [tex]\mathbf A^4[/tex] has a corresponding eigenvalue of 4096.

[tex]\mathbf A^{-1}\vec v=\lambda\vec v\implies\mathbf A\cdot\mathbf A^{-1}\vec v=\mathbf A\cdot\lambda \vec v\implies\vec v=8\lambda\vec v[/tex]

For this to hold, we require [tex]8\lambda=1[/tex], or [tex]\lambda=\frac18[/tex]. So [tex]\mathbf A^{-1}[/tex] has a corresonding eigenvalue of 1/8.

[tex](\mathbf A+4\mathbf I_n)\vec v=\mathbf A\vec v+4\mathbf I_n\vec v=8\vec v+4\vec v=12\vec v[/tex]

so that [tex]\mathbf A+4\mathbf I_n[/tex] has an associated eigenvalue of 12.

[tex]5\mathbf A\vec v=5\cdot8\vec v=40\vec v[/tex]

so the eigenvalue of [tex]5\mathbf A[/tex] is 40.

The matrix [tex]A^{4}[/tex]   has eigen value of 4096.

The matrix [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is 40.

Eigenvalues are the set of scalar values that is associated with the set of linear equations most probably in the matrix equations.

The eigenvectors are also termed as characteristic roots.

If v is an eigen vector of matrix A associated with eigen value [tex]\lambda[/tex].

   Then,         [tex]Av=\lambda v[/tex]

it is given that, v is an eigen vector of A with eign value 8.

so taht,  [tex]Av=8v[/tex]

We have to find eigen value for [tex]A^{4}[/tex].

 [tex]A^{4}v=A^{3}(8v)=8^{4}v=4096v[/tex]

Thus, [tex]A^{4}[/tex] has eigen value of 4096.

If A has eigen value of 8 . Then [tex]A^{-1}[/tex] has eigen value of [tex]\frac{1}{8}[/tex]

The eigen value of (A+4I) is computed as,

        [tex](A+4I_{n})v=Av+4v=8v+4v=12v[/tex]

The eigen value of (A+4I) is 12.

The eigen value of [tex]5Av[/tex] is,

       [tex]5Av=5*8v=40v[/tex]

The eigen value of [tex]5Av[/tex] is 40.

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Answer:

a) The mean of the sampling distribution of p is 0.4.

The standard deviation of the sampling distribution of p is 0.0538.

b) The shape of the sampling distribution of p is approximately normal because the sample size is large.

c) z=0.19

Step-by-step explanation:

We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.

a) The mean of the sampling distribution is equal to the mean of the distribution p:

[tex]\mu_p=p=0.4[/tex]

The standard deviation of the sampling distribution is:

[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538[/tex]

b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.

c) The z-score fof p-0.41 can be calculated as:

[tex]z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19[/tex]

The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.

From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.

The standard deviation will be:

= [(✓0.4 × ✓0.6) / ✓83)]

= 0.0538

Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.

The standard normal z-score will be:

= (0.41 - 0.4) / 0.0538

= 0.10/0.0538

= 0.19.

Therefore, the standard normal z-score is 0.19.

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Answer:

I don't know you're options but the answer could be 30.

Answer:

Contribution

Step-by-step explanation:

The verb would be used as:

I like to contribute to discussions.

The noun would be used as:

My contribution to the discussion was alright.

I hope this helped! (Sorry for the dry examples, I couldn't think of anything else)

Answer:

2

Step-by-step explanation:

Answer: 9?

Step-by-step explanation: I think.

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the population proportion of gamers that prefer consoles is less than 28% as the manufacturer claims.

Of 341 surveyed players, 89 said that they prefer using a console.

The sample resulting sample proportion is p'= 89/341= 0.26

If the company claims is true then p<0.28, this will be the alternative hypothesis of the test.

H₀: p ≥ 0.28

H₁: p < 0.28

α: 0.05

To study the population proportion you have to use the approximation of the standard normal [tex]Z= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]≈N(0;1)

[tex]Z_{H_0}= \frac{0.26-0.28}{\sqrt{\frac{0.28*0.72}{341} } }= -0.82[/tex]

This test is one-tailed left, i.e. that you'll reject the null hypothesis to small values of Z, and so is the p-value, you can obtain it looking under the standard normal distribution for the probability of obtaining at most -0.82:

P(Z≤-0.82)= 0.206

Using the p-value approach:

If p-value ≤ α, reject the null hypothesis

If p-value > α, don't reject the null hypothesis

The decision is to not reject the null hypothesis.

Then at a level of 5%, you can conclude that the population proportion of gamers that prefer playing on consoles is at least 28%.

I hope this helps!

Final answer:

The incorrect statement about the normal distribution is that its distribution function generally has a bell shape when graphed. This shape relates to the probability density function, not the cumulative distribution function, which actually has an S-shaped curve.

Explanation:

When we discuss the properties of a normal distribution, we are dealing with a continuous probability distribution that is widely used across many fields. The statement in question is identifying characteristics of the cumulative distribution function (CDF) of a normal distribution. The correct attributes of this function are that it ranges from 0 to 1, it increases as the quantity increases, and it returns the probability that the outcome from the normal distribution is a certain quantity or lower. However, the statement that the distribution function generally has a bell shape when graphed is incorrect regarding the CDF. The bell shape is a characteristic of the probability density function (PDF) and not the CDF, which increases from 0 to 1 in an S-shaped curve.

The standard normal distribution, which is a special case of the normal distribution with a mean (μ) of zero and a standard deviation (σ) of one, is used for a variety of applications in psychology, business, engineering, and other fields. Understanding the properties of the normal distribution — particularly the standard normal distribution — is crucial for interpreting data and using statistical methods.

It is also important to note that the integral of the normal distribution across its entirety (from ∞ to ∞) is equal to one. This area under the curve represents the total probability of all outcomes and justifies why the CDF ranges from 0 at the minimum to 1 at the maximum end of the distribution.

Answer:

Step-by-step explanation:

Find the volume of a sphere 12 yards a crossed

Answer:

73/100

Step-by-step explanation:

.73 is 73 hundredth which is also 73/100.

Answer:

73/100

Step-by-step explanation:

.73

There are 2 numbers after the decimal, which means the denominator is 1 with 2 zeros after it

73/100

This is as simple as it get since 73 is a prime number

Answer:

V = 18.75

Method: First find the area of the pyramid on top, then find the area of the rectangular prism below.

V of the pyramid = 3.75 ft³

V of the rectangular prism = 15 ft³

3.75 + 15 = 18.75

Hope this helped!!  :)

First, find the volume of the pyramid on top:

(2.5 * 3 * 1.5) / 3 = 3.25

Then, find the volume of the rectangular prism:

2.5 * 3 * 2 = 15

Add the two volumes together:

3.25 + 15 = 18.25

To explain: you found the volume of each separate part and added it together.

Answer:

8π yards

Step-by-step explanation:

A circle subtends a total angle of 360 ° from its center.The length of an arc is directly proportional to the angle it subtends from the circle's center. The arc's length can therefore be calculated as:

C=πd=2πr

Where C is circumference, d is diameter and r is radius. Given r as 16 yards then the arc length which is equivalent to circumference is given as

[tex]C=\frac {90}{360}*2*\pi*16=8\pi[/tex]

Answer:

2,144.7 i just answered the other one is wrong

Step-by-step explanation:

If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.

If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.

Traders engage in foreign currency trading to potentially profit from fluctuations in currency values. In this example, the exchange rates are as follows: 1 U.S. dollar buys 0.8167 Euros, and 1 Euro buys 145.8038 yen. We are asked to define two functions: f(x) represents the number of Euros you can buy with x dollars, and g(x) represents the number of yen you can buy with x Euros.

1. To determine the number of Euros you can buy with x dollars, multiply the exchange rate between dollars and Euros by x. In this case, the exchange rate is 0.8167 Euros per dollar. Therefore, the function f(x) is f(x) = 0.8167x.

2. To find the number of yen you can buy with x Euros, multiply the exchange rate between Euros and yen by x. In this case, the exchange rate is 145.8038 yen per Euro. The function g(x) is g(x) = 145.8038x.

For example:

- If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.

- If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.

These functions enable traders to evaluate the quantity of foreign currency they can acquire or exchange based on the prevailing exchange rates.

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Answer:

99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Step-by-step explanation:

We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.

Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.

Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;

                              P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~ [tex]t__n__1+_n__2-2[/tex]

where, [tex]\bar X_1[/tex] = average gold thickness of control-immersion tip plating = 1.5 μm

[tex]\bar X_2[/tex] = average gold thickness of total immersion plating = 1.0 μm

[tex]s_1[/tex] = sample standard deviation of control-immersion tip plating = 0.25 μm

[tex]s_2[/tex] = sample standard deviation of total immersion plating = 0.15 μm

[tex]n_1[/tex] = sample of printed circuit edge connectors plated with control-immersion tip plating = 7

[tex]n_2[/tex] = sample of connectors plated with total immersion plating = 5

Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex]   =  [tex]\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }[/tex]  = 0.216

Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 99% confidence interval for the difference between the mean population mean, ([tex]\mu_1-\mu_2[/tex]) is ;

P(-3.169 < [tex]t_1_0[/tex] < 3.169) = 0.99  {As the critical value of t at 10 degree of

                                              freedom are -3.169 & 3.169 with P = 0.5%}  

P(-3.169 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 3.169) = 0.99

P( [tex]-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99

P( [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99

99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =

[ [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]

= [ [tex](1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] , [tex](1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] ]

= [0.099 μm , 0.901 μm]

Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].

Answer:

The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.

( 0.17971  , 0.82028)

Step-by-step explanation:

Step:-(i)

Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.

Given sample size n₁ = 7

mean of first sample x₁⁻ =1.5 μ m.

Standard deviation of first sample S₁ = 0.25 μ m

Given data  Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.

Given second sample size n₂ = 5

The mean of second sample x⁻₂ =  1.0 μ m

Standard deviation of first sample S₂ = 0.15 μ m

Level of significance ∝ =0.01 or 99%

Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10

tabulated value t = 2.764

Step(ii):-

The 99% of confidence intervals for μ₁- μ₂ is determined by

(x₁⁻ - x⁻₂)  - z₀.₉₉ Se((x₁⁻ - x⁻₂) ,  (x₁⁻ - x⁻₂)  + z₀.₉₉ Se((x₁⁻ - x⁻₂)

where         [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]

                   [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{0.25^2 }{7 } +\frac{0.15^2 }{5 } } = 0.115879[/tex]

[1.5-1.0 -  2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]

(0.5-0.32029,0.5+0.32029

( 0.17971  , 0.82028)

Conclusion:-

The 99% of confidence intervals for μ₁- μ₂ is determined by

( 0.17971  , 0.82028)

Answer:

17,19,23

that the prime number

Answer:

A prime number is a number whose only factors are one and itself

Prime numbers between 15 and 25:

17, 19, 23

1+3+5+7+...+999 =

= 1+2+3+4+...+500

     +1+2+3+...+499

= 2·(1+2+3+...+499) + 500

= 2·(499·500)/2 + 500

= 499·500 + 500

= 500·(499 + 1)

= 500·500

= 250.000

To find the sum of the series 1 + 3 + 5 + 7 + ... + 999, we recognize this is the sum of the first 500 odd numbers. Using the property that the sum of the first n odd numbers is n², we find the sum to be 500², which is 250000.

Summation of Odd Numbers:

To find the sum of the series 1 + 3 + 5 + 7 + ... + 999 using Gauss's method, we first recognize that this sequence is a series of the first 500 odd numbers.

An interesting property to note is that the sum of the first n odd numbers is n².

For example:

→ 1 = 1 (which is 1¹)

→ 1 + 3 = 4 (which is 2²)

→ 1 + 3 + 5 = 9 (which is 3³)

And so on...

In general, the [tex]n_{th[/tex] odd number can be expressed as 2n - 1.

For 999, to find its position in the series, note that 999 is the 500th odd number (since 2*500 - 1 = 999).

Thus, the sum of the first 500 odd numbers is:

→ Sum = 500²

           = 250000

→ So, the sum of the series 1 + 3 + 5 + 7 + ... + 999 is 250000.

Step-by-step explanation:

[tex](7m + 8)(4m + 1) \\ = 7m(4m + 1) + 8(4m + 1) \\ = 28 {m}^{2} + 7m + 32m + 8 \\ \purple { \bold{= 28 {m}^{2} + 39m + 8}}[/tex]

Answer:

You can use foil or the box method.

Step-by-step explanation:

Sorry it looks so bad :P

Answer:

it d (42)

Step-by-step explanation:

Answer:

d: 42

Step-by-step explanation:

Answer and explanation:

Here is code:

credit = int(input("Enter credits : "))

if credit < 7:

print("You are a Freshman")

elif credit >= 7 and credit < 16:

print("You are a Sophomore")

elif credit >= 16 and credit < 26:

print("You are a Junior")

elif credit >= 26:

print("You are a Senior")

Output: check image

To calculate a student's class standing, use conditional statements to determine the appropriate class based on the number of credits earned.

To write a program that calculates a student's class standing based on the number of credits earned, you can use conditional statements. Here's a step-by-step explanation:

Take the input of the number of credits earned from the user.

Use conditional statements (if, else if) to check the number of credits and assign the appropriate class standing.

If the number of credits is less than 7, then the student is a Freshman. If it is at least 7 and less than 16, then the student is a Sophomore.

If it is at least 16 and less than 26, then the student is a Junior. Otherwise, the student is a Senior.

Display the class standing to the user.

Here's an example pseudocode:

credits = input('Enter the number of credits earned: ')

if credits < 7:

   print('Class Standing: Freshman')

elif credits < 16:

   print('Class Standing: Sophomore')

elif credits < 26:

   print('Class Standing: Junior')

else:

   print('Class Standing: Senior')

Learn more about Calculating class standing here:

https://brainly.com/question/31437017

#SPJ6

[tex]\text{Solve the equation to make it into slope intercept form}\\\\\text{Slope intercept form: y = mx + b}\\\\\text{Solve:}\\\\-2y=6(2x-2)\\\\\text{Distribute the 6 to the variables inside the parenthesis}\\\\-2y=12x-12\\\\\text{Divide both sides by -2}\\\\y=-6x+6\\\\\text{The slope would be -6 and the y intercept would be 6}\\\\\boxed{\text{Slope: -6 y-intercept: 6}}[/tex]

Answer:

He has 5 children

Step-by-step explanation:One boy is a brother to all of them

A number is an arithmetic value that is used to represent a quantity and calculate it. The number of children that Mr. Smith will have is 5.

A number is an arithmetic value that is used to represent a quantity and calculate it. Numericals are written symbols that represent numbers, such as "3."

As it is mentioned that Mr. Smith had 4 daughters and each of the daughters had a brother. Now, if Mr. Smith had just a single son then he will be a brother to every daughter. Therefore, the number of children that Mr. Smith will have is 5.

Learn more about Numbers:

https://brainly.com/question/17429689

Answer:the answer is (A)

Step-by-step explanation:

Because the survey face-to-face, one at a time, with an administrator of the school the students would likely reply that they do not text and drive even if they actually do. This response bias would cause the sample proportion to underestimate the value of the true proportion of all students at this school that text and drive.

The design of the survey contains response bias, which is likely to underestimate the true proportion of students who text while driving due to students' reluctance to admit to prohibited and risky behaviors to an authority figure.

The potential bias present in the design of this survey is response bias.

This type of bias occurs when the way a question is asked leads the respondents to give a certain answer

. In this case, by directly asking the students if they engage in the prohibited behavior of texting while driving, there is a likelihood that students may not answer truthfully due to fear of repercussions, the desire to give socially acceptable answers, or reluctance to admit to risky behaviors in a face-to-face setting with a school authority figure.

The likely direction of the bias is that it would underestimate the true proportion of students who text while driving since students might not want to admit to a behavior that is both dangerous and frowned upon while talking to a school administrator.