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melomori [17]
3 years ago
8

A shirt originally cost $42.54, but it is on sale for $29.78. What is the percentage decrease of the price of the shirt? If nece

ssary, round to the nearest percent.
Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

The difference is 30%

Step-by-step explanation:

To find the percent decrease, start by finding the difference in costs.

42.54 - 29.78 = 12.76

Now divide that amount by the original cost.

12.76/42.54 = 30%

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Alik [6]

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3 years ago
A certain bacteria culture grows at a rate of 18% everyday. There are currently 420 bacteria. How many bacteria will there be on
spayn [35]

In this question, it is given that

A certain bacteria culture grows at a rate of 18% everyday. There are currently 420 bacteria.

And we have to find number of bacterias on day *.

For that we use the following formula

N(t) =N_{0} (1 +r)^t

N(0)= 420, r = 0.18 , and \ t = 8days

Substituting these values in the formula, we will get

N(8)= 420(1+0.18)^8
\\
N(8)=1579

THerefore on day 8, number of bacterias are 1579.

4 0
3 years ago
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Find the x- and y-intercepts of the graph of 2x - y = 24. State your answers as
Ghella [55]
(12,-24)
x intercept is 12 & y intercept is -24
6 0
3 years ago
Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A
lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let n(A) the number of elements in A.

Remember, the number of elements in A_1 \cup A_2 \cup A_3 satisfies

n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)

Then,

a) If A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200, and if A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000

Since A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1

So

n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600

b) Since the sets are pairwise disjoint

n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193

8 0
3 years ago
What does 3/8 + 7/8=
ValentinkaMS [17]

Answer:

since the denominators are the same we add 3 and 7 in the numerator to get 10/8 or 1 2/8, and we can simplify to 5/4 or 1 1/4.

Hope this helps and brainliest please

5 0
3 years ago
Read 2 more answers
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