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3 years ago

7

Gary is reaching into a bag that contains a red ball a blue ball and a white ball. at the same time eric is tossing a coin into

the air. what is the probability of gary pulling a white ball at the same time as erics coon landing on heads
a. 1/8
b. 1/5
c. 1/7
d. 1/6

1 answer:

il63 [147K]3 years ago

6 0

Answer:

A:1/8

Step-by-step explanation:

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3 years ago

The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or

tigry1 [53]

<u>Answer-</u>

<em>The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%</em>

<u>Solution-</u>

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

$P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}$

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}$

$P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67$

Putting these values,

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{0.15}{0.67} =0.224=22.4\%$

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4 years ago

PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)

ExtremeBDS [4]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 3 &,& -4~) 
% (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}$

$\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23$

side note: multiplying by the LCD of both sides is just to get rid of the denominators

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3 years ago