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3 years ago

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25 Points! Please answer asap! Carly stated “All pairs of rectangles are dilations”. Which pair of rectangles would prove that C

arly’s statement is incorrect? (Images below)

2 answers:

elena55 [62]3 years ago

8 0

Answer:

C

Step-by-step explanation:

A. First two rectangles are dilations because

$\dfrac{2}{4}=\dfrac{4}{8}=0.5$

B. Second two rectangles are dilations because

$\dfrac{2}{4}=\dfrac{3}{6}=0.5$

C. Third two rectangles are not dilations because

$\dfrac{3}{4}\neq \dfrac{2}{3}$

D. Fourth two rectangles are dilations because

$\dfrac{3}{4}=\dfrac{1.5}{2}=0.75$

skad [1K]3 years ago

5 0

Answer: C

I just did the pre test

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Plzz anyone solve all answers plzzzzzzzz

algol13

You posted a lot of problems here. In the future please only post one problem at a time. Thank you.

I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.

==========================================

Problem 1

{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).

This means we have some number d added on to each term to get the next term.

first term = 18

second term = first term + d = 18+d = a

third term = second term + d = (18+d)+d = 18+2d = b

fourth term = third term + d = (18+2d)+d = 18+3d = -3

----

Let's solve that last equation for d

18+3d = -3

18+3d-18 = -3-18

3d = -21

3d/3 = -21/3

d = -7

----

The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term

first term = 18

second term = first term + d = 18+d = 18+(-7) = 18-7 = 11

third term = second term + d = 11+d = 11+(-7) = 11-7 = 4

fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3

----

We see that a = 11 and b = 4 are the second and third terms respectively.

Therefore, a+b = 11+4 = 15

-------------

<h3>Answer: 15</h3>

==========================================

Problem 2

A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.

We want to know which values of 4*n are between 10 and 250.

----

Divide both 10 and 250 by 4 to get the following

10/4 = 2.5

250/4 = 62.5

If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.

If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62

----

The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"

You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below

{a, a+1, a+2, ..., b-2, b-1, b}

where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula

c = b-a+1

will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}

----

In this case, a = 3 and b = 62, making

c = b-a+1

c = 62-3+1

c = 60

There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}

There are 60 multiples of four that are between 10 and 250.

-------------

<h3>Answer: 60</h3>

4 0

3 years ago

Cho hệ đầy đủ ba biến cố {A,B,C} với 2P(A)=P(B)=2P(C). Biết biến cố F thỏa mãn:

ExtremeBDS [4]

267272718;27277282,727272?2727

4 0

3 years ago

What percent is shaded in the grid? Answer:

Keith_Richards [23]

I can’t see a picture of the grid

4 0

3 years ago

(3a ^2 + 1) - ( 4+2a^2)

xz_007 [3.2K]

Answer:

a^2 - 3

Step-by-step explanation:

(3a^2 + 1) - (4 + 2a^2)

3a^2 + 1 - 4 - 2a^2

a^2 - 3

8 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