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.