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3 years ago

Carol ate 2/5 of the cake. Dima ate 3/5 of the REMAINING CAKE, mom ate the rest.

1 answer:

5 0

"<span>How many times more did Carol eat compared to Dima?" is quite different from the previous question, in which you asked "how much more did Carol eat than Dima?" I think they ate 10/25 and 9 /25 of the cake, respectively.

10/25 is greater than 9/25, obviously. We need to calculate "how many times greater is 10/25 than 9/25? So we divide 9/25 INTO 10/25. This comes out to 10/9, or 1.11 times as much. Carol eats 11% more cake than Dima. Hope she doesn't gain too much extra weight. ;)</span>

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A Nevada roulette wheel had 38 pockets, labeled as 0 00 1 2 3 ... 35 36. One bet is single number. If you bet $1 on a number (an

slavikrds [6]

Answer:

-0.0526

Step-by-step explanation:

Let X be the random variable denoting the net gain(in dollars) for a single trial(one bet).

Assuming that each number in the wheel is equally likely, probability of the outcome being a victory is $\frac{1}{38}$ and probability of failure is $\frac{37}{38}$ . For a win, X takes value 35 and for a loss X takes value -1. So the model is,

$P(X=35) = \frac{1}{38}$

$P(X=-1) = \frac{37}{38}$

$P(X=i) = 0 \;if\; i \neq 38, i\neq -1$

The mean for one bet is $E(X) = \sum xP(X=x) = 35\times\frac{1}{38} - 1\times\frac{37}{38} = -0.0526$

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3 years ago

What fraction of an hour is 42 minutes? Give your answer in its simplest form

lorasvet [3.4K]

$\frac{60}{42} = \frac{10}{7}$

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3 years ago

1 Pound brown sugar is equivalent to _____ cups

iogann1982 [59]

Answer:

3 1/2 or said as 3.5 cups is equal to 1 pound

Step-by-step explanation:

1 Pound brown sugar is equivalent to ___3 1/2 or 3.5 __ cups

Hope This Helped

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2 years ago

1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i

Katena32 [7]

Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:

$f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}$

Answer B.

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3 years ago

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Here you go! I hope I can get brainliest!

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3 years ago