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3 years ago

What percent of the model is shaded?

Mathematics

2 answers:

maria [59]3 years ago

7 0

Answer:

1 is 73% 2 is 73/100 3 is 0.73 and 4 is 73/100

Step-by-step explanation:

i took the quiz

Burka [1]3 years ago

5 0

Answer:

i just did it so here.

hope this helps and have a great day:).

You might be interested in

In a recent year, 31% of all college students were enrolled part-time. If 8.6 million college students were enrolled part-time t

Ronch [10]

Answer:

28 million college students.

Step-by-step explanation:

We can represent all college students as x. This means that all part-time college students are $\frac{31}{100}x$ . Because all part-time college students are also equal to 8.6 million, we have $8,600,000=\frac{31}{100}x$ . To find the number of total college students, or x, we have to multiply by 100/31 to get rid of the fraction on the other side:

$8,600,00\cdot\frac{100}{31}=x$

The left side of the equation evaluates as $27,741,935.5$ . Because we are asked to round to the nearest million, we have to round to 28 million, which is the closest.

5 0

2 years ago

Sara is mixing together a fruit punch for a party. She's made 4 gallons of punch with a mixture of 50% juice. Her mother tells h

Nata [24]

$\bf \begin{array}{lccclll}
&amount(gallons)&juice&\textit{juice amount}\\
&--------&-----&-----\\
\textit{50\% punch}&4&0.50&(4)(0.50)\\
\textit{pure juice}&x&1.00&(x)(1.00)\\
-----&-----&-----&-----\\
mixture&4+x&0.60&(4+x)(0.60)
\end{array}$

notice, that, pure juice is 100% juice, dohhh, thus 100/100 = 1.00
50% is 50/100 or 0.50 in decimal format

so..... whatever those two quantities amount to, that is, the 50% and pure juice, or (4)(0.50) + (x)(1.00)
they will equal the mixture desired 60% juice, or 0.60, namely (4+x)(0.60)

thus (4)(0.50) + (x)(1.00) = (4+x)(0.60)

solve for "x"

4 0

3 years ago

In the rolling of two fair dice calculate the following: P(Sum of the two dice is 7) = ______

vesna_86 [32]

Answer:

P(Sum of the two dice is 7) = 6/36

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

A fair dice can have any value between 1 and 6 with equal probability. There are two fair dices, so we have the following possible outcomes.

Possible outcomes

(first rolling, second rolling)

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)

(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)

(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)

(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

There are 36 possible outcomes.

Desired outcomes

Sum is 7, so

(1,6), (6,1), (5,2), (2,5), (3,4), (4,3).

There are 6 desired outcomes, that is, the number of outcomes in which the sum of the two dice is 7.

Answer

P(Sum of the two dice is 7) = 6/36

7 0

4 years ago

A batch consists of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are rando

Allushta [10]

<u>Answer:</u>

The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744

<u>Solution:</u>

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) $\times$ p(getting 2 good coils for second selection)