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3 years ago

I need to find out the pythagorean theorem plss HELLPP NOWWWWW!!!!!!!!

Mathematics

2 answers:

QveST [7]3 years ago

8 0

Pythagorean theorem is a^2+ b^2=c^2 so just plug in the numbers

Pavlova-9 [17]3 years ago

3 0

Answer:

find the square value of side b.

find the square value of side c.

Subtract b^2 from c^2.

Find the root square value of the difference is the value of a.

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Need help asap please

Trava [24]

Answer:

$\displaystyle no$

Step-by-step explanation:

The triangle on the right is bigger than the triangle on the left.

I am joyous to assist you at any time.

3 0

3 years ago

Identify the unit rate in the graph. A) $5.50 B) $11.00 C) $22.00 D) $27.50

Lera25 [3.4K]

A) $5.50 Is correct.

7 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

Can Someone Help Me With These Three Please?

lakkis [162]

Answer:

ΔHJK shows G as the orthocenter
Show A.F is the median of BC
18

Step-by-step explanation:

1) This is a vocabulary question. The orthocenter is the point where the altitudes intersect. Of course, each altitude is a segment from a vertex that is perpendicular to the opposite side of the triangle.

Perhaps it could be useful to remember the prefix ortho- means perpendicular, as in orthogonal. Each altitude is perpendicular to one of the sides of the triangle.

2) To do this proof, you would find midpoints of segments AB and AC, write the equations of the lines through them from the opposite vertex, then solve those equations to find point G. You would then write the equation for line AG and find its intersection point with segment BC (point F). The last step is to show that point F is the midpoint of BC. (It might be easier to show that midpoint F is on line AG.)

The closest answer choice, though poorly worded, is the last one: show A.F is the median of BC.

(Strictly speaking, a line segment (A.F) is not a median of a line segment (BC), but can be a median of a triangle, or a bisector of a line segment.)

3) The centroid divides the median into parts with the ratio 2:1. That is, the shorter part differs from the longer one by 2-1 = 1 ratio unit. If those parts differ in length by 6 measurement units, then one ratio unit must be 6 measurement units. The total length of the median is 2+1 = 3 ratio units, or 3×6 measurement units = 18 measurement units.

_____

Comment on A.F

Brainly thinks the name of the segment starting with A and ending with F is a "bad word" so won't let it be posted.

6 0

4 years ago

Z=a + m, for a Help me pls solve for a

gogolik [260]

Its 76-9=^742+8477
= 8531

5 0

4 years ago

Read 2 more answers