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2 years ago

Need help with this one also

Mathematics

1 answer:

andriy [413]2 years ago

4 0

Answer:

scale factor 3 is the answer.

Step-by-step explanation:

Length = 18 in

Breadth = 3 in

height = 10 in

Volume of cuboid = l × b × h

V = 18 × 3 × 10

V = 540 in³

540 in³ is the original volume. The volume needs to be tripled, so multiply the original volume with 3

V = 540 × 3 = 1620 in³

∴ 1620 in³ will be the new volume if it's tripled and the scale factor is 3.

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The type of transformation PQRS undergoes is a . If vertex Q is at (-4, -5), then vertex Q′ is at ?

MissTica

Answer:

(-4, -5) it gives the answer in the question

Step-by-step explanation:

8 0

2 years ago

20 POINTS HELPPPPP

kiruha [24]

For this case we have the following expression:

$\frac {216 ^ {n-2}} {(\frac {1} {36}) ^ {3n}} = 216$

We multiply both sides by: $(\frac {1} {36}) ^ {3n}$

$216 ^ {n-2} = 216 * (\frac {1} {36}) ^ {3n}$

We divide both sides by 216:

$\frac {216 ^ {n-2}} {216} = (\frac {1} {36}) ^ {3n}$

To divide powers of the same base, we place the same base and subtract the exponents:

$216 ^ {n-2-1} = (\frac {1} {36}) ^ {3n}\\216 ^ {n-3} = (\frac {1} {36}) ^ {3n}$

Rewriting:

$(6 ^ 3) ^ {n-3} = (\frac {1} {6 ^ 2}) ^ {3n}\\6 ^ {3n-9} = \frac {1} {6 ^ {6n}}\\6^{ 3n-9} * 6^{ 6n} = 1$

To multiply powers of the same base, we place the same base and add the exponents:

$6^{ 3n-9 + 6n} = 1\\6^{ 9n-9} = 1$

We know that any number raised to zero is 1, $a ^ 0 = 1.$

So, for equality to be true:

$9n-9 = 0\\9n = 9\\n = \frac {9} {9}\\n = 1$

Answer:

$n = 1$

3 0

2 years ago

Read 2 more answers

Seventh grade

alexandr402 [8]

Answer:

$690.86

Step-by-step explanation:

100% from the original cost is: 2 x 863.57 = $1,727.14

40/100 = X/1727.14 (40 percent is what of 1,727.40) (%/100 = is/of)

(40 * 1,727.14) / 100 = $ 690.86

6 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

2 years ago

What is the solution set of x for the given equation? x^2/3=x^1/3+4=6 A. -2, -1 B. 2, -1 C. -8, -1 D. 8, -1 E. 2, 8

Valentin [98]

Answer:

Interpreting as: x^2/3=x^1/3+4=6 A

Input:

x^2/3 = x^(1/3) + 4 = 6 A

3 0

2 years ago