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3 years ago

7

Yuron is sketching a scale drawing of a famous painting. The painting is 54 inches wide and 72 inches long. He used a scale fact

or of 2/3 to make his sketch. How many inches wide is Yuron’s sketch?

1 answer:

arsen [322]3 years ago

5 0

Answer:54*(2/3)= 36= wide

72*(2/3)= 48= long

Step-by-step explanation:

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Suppose that for a certain experiment p(b)equals=0.5 and p(avertical line upper bb)equals=0.2. find p(aintersect∩b).

raketka [301]

P(a | b) = p(a ∩ b)/p(b) . . . . . a definition worth remembering
0.2 = p(a ∩ b)/0.5 . . . . . . . . fill in the given values
0.2*0.5 = 0.1 = p(a ∩ b)

p(a ∩ b) = 0.1

5 0

3 years ago

What does 108 × 1/3 equal?

Alisiya [41]

Answer:

It would be 36

Explanation:

<em>First:</em>

<em>Convert any mixed numbers to fractions.</em>

<em>Then your initial equation becomes:</em>

<em>108/1 ×1/3</em>

<em> </em>

<em>Applying the fractions formula for multiplication,</em>

<em>=108×1/1×3 = 108/3</em>

<em> </em>

<em>Simplifying 108/3, the answer is</em>

<em>=36</em>

hope it helped. :)

5 0

3 years ago

Use the Pythagorean Theorem to determine if a

MakcuM [25]

Answer:

how can it be triangle with 0ft length

6 0

3 years ago

use the explicit formula an = a1 + (n - 1) * d to find the 500th term of the sequence below. 24,30,36,42,48,... A. 3018 B. 3042

marishachu [46]

Answer:

Option A.

Step-by-step explanation:

The given sequence is

24, 30, 36, 42, 48, ...

It is an AP. Here,

First term = 24

Common difference = 30-24 = 6

The given explicit formula for nth term is

$a_n=a_1+(n-1)d$

where, $a_1$ is first term, d is common difference.

Substitute $a_1=24, d=6 \text{ and }n=500$ in the above formula.

$a_{500}=24+(500-1)(6)$

$a_{500}=24+(499)(6)$

$a_{500}=24+2994$

$a_{500}=3018$

The 500th term of the sequence is 3018.

Therefore, the correct option is A.

3 0

3 years ago

If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumpe

lakkis [162]

If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper

Answer: We are given there are 49 cars

Also these 49 cars are in a line stretched 528 feet.

Now the average length of the car is:

Average length = $\frac{(length-covered-by-49-cars)}{Number-of-cars}$

= $\frac{528}{49}=10.78$

Therefore, the average length of car is 10.78 feet

6 0

3 years ago