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2 years ago

What is 76 times 468

Mathematics

2 answers:

Mariana [72]2 years ago

7 0

The answer is 35,568.

r-ruslan [8.4K]2 years ago

4 0

The answer to your question is 35,568

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Write an expression using the distributive property to find the product of 7×63

Nataliya [291]

7*(60+3)=7*60+7*3= 420+21=441

4 0

3 years ago

Help me with this question please?

UNO [17]

$\bf 3^{n+2}-3^n=8\cdot 3^n\implies \boxed{3^n\cdot 3^2}-3^n=8\cdot 3^n
\\\\\\
\stackrel{common~factor}{3^n}(3^2-1)=8\cdot 3^n\implies 3^2-1=\cfrac{8\cdot \underline{3^n}}{\underline{3^n}}\implies 3^2-1=8
\\\\\\
9-1=8\implies 8=8$

6 0

2 years ago

What is the domain of the relation y = arccscx?

Stella [2.4K]

Answer:

$y=arc\csc x$

Step-by-step explanation:

The given function is $y=arc\csc x$ .

The domain refers to all values of x for which this function is defined.

Recall that: the domain of $y=arc \sin (x)$ is $-1\le x\le1$

And we know $y=arc\csc x$ is the reciprocal of $y=arc \sin (x)$ .

Therefore the complement of the domain of $y=arc \sin (x)$ which is $(-\infty,-1]\cup [1,+\infty)$ is the domain of $y=arc\csc x$

5 0

2 years ago

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What

galina1969 [7]

Answer:

Part A:

The probability that all of the balls selected are white:

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

$P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

$P(D_i)=\frac{1}{6}$ for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

$P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}$

$P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}$

$P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}$

$P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}$

$P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}$

$P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0$

Part A:

The probability that all of the balls selected are white:

$P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)$

$P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\ P(A)=\frac{5}{66}=0.075757576$

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find $P(D_3|A)$

The data required is calculated above:

$P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516$

7 0

3 years ago

7b + 15 +4 it says write two equivalent expressions.Help

VARVARA [1.3K]

Answer:

1. 4b+3b+15+4

2. 7b+19

Step-by-step explanation:

Both are equivalent expressions :))

7 0

2 years ago