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2 years ago

Help me pls it’s urgent

Mathematics

1 answer:

earnstyle [38]2 years ago

3 0

Answer:

<A = 115

<C = 50

Step-by-step explanation:

<A + <B + <C = 180

(4X -13) +15 +(X+18) =180

5X +20 =180

5X = 160

X= 32

<C = 32+18 = 50

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Find the y-intercept for the equation y=15x-3

Anestetic [448]

Answer:

(0,-3)

Step-by-step explanation:

the y intercept is the last number of the expression; in this case it's negative 3. x is always 0

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2 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.

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2 years ago

What ratio is equivalent to the ratio 14:7 A 1:7 B2:7 C2:1

svp [43]

Answer:

c 2:1

Step-by-step explanation:

If you simplify 14:7 it is 2:1 (divide by 7)

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2 years ago

Read 2 more answers

Please help it’s a test and I can’t get logged out

HACTEHA [7]

Answer:

the anwer is B ( i mean second option)

And you can try it

you will find ;

$y = \frac{x}{3} - 1$

HAVE A NİCE DAY

Step-by-step explanation:

GREETİNGS FROM TURKEY ツ

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2 years ago

Find the angle between the given vectors to the nearest tenth of a degree. u = , v = (2 points)

Liono4ka [1.6K]

Answer:

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

we will be using the formula below to calculate the angle between the two vectors;

$u*v = |u||v| cos \theta$

$\theta$ is the angle between the two vectors.

u = 8i + 7j and v = 9i+7j

u*v = (8i + 7j )*(9i + 7j )

u*v = 8(9) + 7(7)

u*v = 72+49

u*v = 121

|u| = √8²+7²

|u| = √64+49

|u| = √113

|v| = √9²+7²

|v| = √81+49

|v| = √130

Substituting the values into the formula;

121= √113*√130 cos θ

cos θ = 121/121.20

cos θ = 0.998

θ = cos⁻¹0.998

θ = 3.6° (to nearest tenth)

Hence, the angle between the given vectors is 3.6°

5 0

2 years ago