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harkovskaia [24]
3 years ago
7

Will mark brainiest for the first one with the correct answer.

Mathematics
2 answers:
just olya [345]3 years ago
5 0
The probably of it being coconut or kiwi is 14 over 15
Ber [7]3 years ago
4 0

Answer:

for pulling out a killer kiwi it would be 5/15 and 9/15 for crazy coconut

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Given the volume of Figure A is 512cm ^3and Figure B is 343cm^3, find the ratio of the perimeter from Figure A to Figure B.
sp2606 [1]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{\textit{figure A}}{\textit{figure B}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{512}}{\sqrt[3]{343}}\qquad \begin{cases} 512=&2^9\\ &2^{3\cdot 3}\\ &(2^3)^3\\ 343=&7^3 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt[3]{(2^3)^3}}{\sqrt[3]{7^3}} \\\\\\ \cfrac{s}{s}=\cfrac{2^3}{7}\implies \cfrac{s}{s}=\cfrac{8}{7}\implies s:s = 8:7\impliedby \textit{ratio of the }\stackrel{sides~and}{perimeters}

5 0
3 years ago
Pls help 20 points TYYY
Mumz [18]
Answer: A

Explanation:
It is exponential because the y value is multiplied by 3 each time.
6 0
3 years ago
What percent of 8 is 14
seraphim [82]

Answer:

57.14

Step-by-step explanation:

8/14 =x/100

cross multiply

800= 14x

x=57.1428

7 0
3 years ago
Read 2 more answers
A simple random sample of size nequals57 is obtained from a population with muequals69 and sigmaequals2. Does the population nee
dusya [7]

Answer:

The population does not need to be normally distributed for the sampling distribution of \bar{X} to be approximately normally distributed. Because of the central limit theorem. The sampling distribution of \bar{X} is approximately normal.

Step-by-step explanation:

We have a random sample of size n = 57 from a population with \mu = 69 and \sigma = 2. Because n is large enough (i.e., n > 30) and \mu and \sigma are both finite, we can apply the central limit theorem that tell us that the sampling distribution of \bar{X} is approximatelly normally distributed, this independently of the distribution of the random sample. \bar{X} is asymptotically normally distributed is another way to state this.

6 0
3 years ago
Sarah invested $1000 in an account paying 5.5% interest compounded semi-annually. How long will it take for the account balance
NISA [10]

Answer: it will take 43.25 years

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $1000

A = $10450

r = 5.5% = 5.5/100 = 0.055

n = 2 because it was compounded 2 times in a year.

Therefore,.

10450 = 1000(1 + 0.055/2)^2 × t

10450/1000 = (1 + 0.0275)^2t

10.45 = (1.0275)^2t

Taking log of both sides, it becomes

Log 10.45 = 2t log 1.0275

1.019 = 2t × 0.01178 = 0.02356t

t = 1.019/0.02356

t = 43.25 years

4 0
3 years ago
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