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1 year ago

2(x−8)=x+5x x= −4 x= −16 x= −2 x = 0

Mathematics

2 answers:

olasank [31]1 year ago

8 0

Answer: X = -4

Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.

Margarita [4]1 year ago

6 0

Correct answer: x= -4
Math-way is a great help for questions like these.

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Carol and Jan leave from the same place and travel on the same road going the same direction. Carol walks at a rate of 2 miles p

kramer

Let, the time when <span>Jan cough up to Carol = x

We know, Distance = Speed * Time.

So, Equation would be: 2x + 5*2 = 6x

2x + 10 = 6x

6x - 2x = 10

4x = 10

x = 10/4 = 2.5

Distance traveled in that time = 2.5 * 6 = 15 miles

In short, Your Answer would be: 15 miles

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3 years ago

I got so many points, so i wanna give some away!!

MrRissso [65]

Answer:

10 thanks by the way

Step-by-step explanation:

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3 years ago

Read 2 more answers

find a parameterization for a circle of radius 3 with center (-4,4,1) in a plane parallel to the yz plane. write your parameteri

madam [21]

Answer:

Following are the solution to the question:

Step-by-step explanation:

$Radius= 3\\\\center= (-4,4,1)\\\\x(t)=-4\\\\y(t)=4+3 \cos \ t\\\\z(t)= 1+ 3 \sin \ t$

6 0

3 years ago

A reporter for a student newspaper is writing an article on the cost of off-campus housing. A sample was selected of 10 one-bedr

olga55 [171]

Answer:

(507.05, 592.95)

Step-by-step explanation:

Given data:

sample mean = $550, sample standard deviation S = $60.05

95% confidence interval , n = 10

For 95% confidence interval for the mean

mean ± M.E.

where M.E. is margin of error = $t_{n-1}, \alpha/2\times\frac{S}{\sqrt{n} }$

Substituting the values in above equation

$=t_{10-1}, 0.05/2\times\frac{60.05}{\sqrt{10} }$

= 2.62×18.99

=42.955

= 550±42.95

=(507.05, 592.95)

6 0

3 years ago

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

​2(x−8)=x+5x x= −4​ ​x= −16​ ​x= −2​ x = 0

2(x−8)=x+5x x= −4 x= −16 x= −2 x = 0