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3 years ago

PLEASE HELP!!

2 answers:

7 0

Answer: You could move Circle 2 10 units to the right and up 5 up.

Doing this would put the centers of the circles at the same location. They would both be at the point (8, 5).

For the dilation, the radius changed from 2 to 6. Solve the following equation.
2x = 6
x = 3

lora16 [44]3 years ago

6 0

Answer: Its actully 4 units up but the rest is correct

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. The ratio of the area of circle A to the area of circle B is 12/5. What is the ratio of their circumferences?

N76 [4]

Answer:

1.5

Step-by-step explanation:

a : b = 1 : 12/5

Area of a circle = πr²

Where : = π = pi = 22/7

R = radius

Let the radius of circle a = 1

Area of circle A = 1π

Radius of circle A = square root of 12/5 = 1.5

The circumference of a circle = 2nr

n = 3.14

circumference of circle A = 2 x n x 1 = 2n

circumference of circle B = 2 x n x 1.5 = 3n

ratio of circumference = 3n / 2n = 1.5

4 0

3 years ago

Word problem for 3(23)

Flauer [41]

This shows the multiplication between 3 and 23. So your word problem will have to show that. Jean buys three packs of gum. Each pack has 23 pieces. How many pieces of gum does Jean have in total? :)

7 0

4 years ago

What is the sum of -36 4/9 - (10 2/9) - (18 2/9)

grigory [225]

let's firstly convert the mixed fractions to improper fractions and then add them up.

$\bf \stackrel{mixed}{36\frac{4}{9}}\implies \cfrac{36\cdot 9+4}{9}\implies \stackrel{improper}{\cfrac{329}{9}} \\\\\\ \stackrel{mixed}{10\frac{2}{9}}\implies \cfrac{10\cdot 9+2}{9}\implies \stackrel{improper}{\cfrac{92}{9}}~\hfill \stackrel{mixed}{18\frac{2}{9}}\implies \cfrac{18\cdot 9+2}{9}\implies \stackrel{improper}{\cfrac{164}{9}} \\\\[-0.35em] ~\dotfill$

$\bf -\cfrac{329}{9}-\cfrac{92}{9}-\cfrac{164}{9}\implies \stackrel{\textit{using the LCD of 9}}{\cfrac{-329-92-164}{9}}\implies -\cfrac{585}{9}\implies -65$

8 0

3 years ago

You’ve bought a half-dozen (six) eggs from the store but you forgot to check them first! The probability that no eggs are broken

GREYUIT [131]

Answer:

$P(X = 0) = C_{6,0}.(0.1416)^{0}.(0.8584)^{6} = 0.4$

$P(X = 1) = C_{6,1}.(0.1416)^{1}.(8584)^{5} = 0.3960$

$P(X = 2) = C_{6,2}.(0.1416)^{2}.(8584)^{4} = 0.1633$

$P(X = 3) = C_{6,3}.(0.1416)^{3}.(8584)^{3} = 0.0359$

$P(X = 4) = C_{6,4}.(0.1416)^{4}.(8584)^{2} = 0.0044$

$P(X = 5) = C_{6,5}.(0.1416)^{5}.(8584)^{1} = 0.0003$

$P(X = 6) = C_{6,6}.(0.1416)^{6}.(8584)^{0} = 0.00001$

b) 56.77% probability that an even number of eggs is broken.

Expectation: 0.8496

Variance: 0.7293

Step-by-step explanation:

For each egg, there are only two possible outcomes. Either it is broken, or it is not. The probability of an egg being broken is independent from other eggs. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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