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Archy [21]
3 years ago
12

The equation 0.3x = 7 was translated from the problem "_____."

Mathematics
2 answers:
Savatey [412]3 years ago
6 0
What number is 30% of 7
pshichka [43]3 years ago
5 0

Answer:

B) 30% of what number is 7.

Step-by-step explanation:

The given equation is 0.3x = 7

Here "x" is a variable.

0.3x = 7 means 0.3 times of x equals 7.

The variable x is assigned for an unknown number and 0.3 = \frac{30}{100} = 30 %

0.3x = 7

30% of x is 7

Which means 30% of the number(x) is 7.

Therefore, the answer B) 30% of what number is 7.

Note: Variables always assigned to find the unknown numbers.

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Answer:

False, true, true, false, false

Step-by-step explanation:

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2 years ago
There are 460 studets enrolled at Freedom Academy. 40 percent of the students are male. How many students are male?
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184 students are male 
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3 years ago
PLZ HELP!!!
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3 years ago
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Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,
MrRissso [65]

Answer:

4.82\cdot 10^{-4}

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

n-1=51

Therefore, the probability of drawing a three at the 2nd draw is

p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784

Then, at the third draw, the previous 2 cards are not replaced, so there are now

n-2=50

cards in the deck. So, the probability of drawing a jack is

p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08

Therefore, the total probability of drawing an ace, a three and then a jack is:

p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}

4 0
3 years ago
What is the range of the function?
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2, 6, 8) → D

each element in the domain ( - 1, 3, 5) maps to exactly one value in the range (2, 6, 8)


7 0
3 years ago
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