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dlinn [17]
3 years ago
8

PLZ HELP NO WRONG ANSWERS

Mathematics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer :

to find area of the square = 6 to the power 2

to find the area of the rectangle = (5x + 12) * 5x

therefore,

square's area = 36

rectangle's area = 5x*5x + 12 * 5x

= 25x * 60x

Now we have to find the sum of both the areas,

so, 36 + 1500 = 1536

Step-by-step explanation:

look I know my answer is wrong but I just wanted points! thnx for the free points! :)))))))))

NANRI! VANAKKAM!

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Please help me with these 2 questions
STatiana [176]

Answer:

#1.)  x = 7

Step-by-step explanation:

#1.) Here,  AB = BC  because they are tangents that meet at a common endpoint

so  5x + 11 = 3x + 25

2x = 14

x = 7

-----------

3 0
3 years ago
Which statements are true? Select three options. ∠F corresponds to ∠F'. Segment EE' is parallel to segment FF'. The distance fro
Gwar [14]

Answer:

The correct options are (1), (3) and (5).

Step-by-step explanation:

The two triangles are shown below.

The measure of ∠F corresponds to ∠F'.

The distance between the points D and origin is of 9 units.

And the distance between the points D' and origin is 3 units.

Thus, the distance from point D' to the origin is One-third the distance of point D to the origin.

Check for similarity:

\frac{D'F'}{DF}=\frac{D'E'}{DE}=\frac{F'E'}{FE}=\frac{1}{3}

Thus, the △DEF is similar to △D'E'F'.

Thus, the correct options are (1), (3) and (5).

4 0
3 years ago
How do u find the mean
dusya [7]
You add all the numbers & divide it by how many numbers 8 +9+9+9+10+10
7 0
3 years ago
Read 2 more answers
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
I need help ASAP giving brainliest!!
Olin [163]

Answer:

<em>isosce</em><em>les</em><em> </em><em>and</em><em> </em><em>right</em>

<em>I'm not sure</em>

3 0
3 years ago
Read 2 more answers
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