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forsale [732]
3 years ago
15

A jar contains 4 red and 8 blue marbles a marble is drawn from the jar what is the probability that a red marble is drawn ?

Mathematics
1 answer:
frosja888 [35]3 years ago
5 0

There is a 1/3 chance that a red marble would be chosen.

To find out you put add up the marbles, so they will make 12 marbles.

Since there are 4 red you put 4/12.

4/12 simplified is 1/3.

So the chance of a red marble being picked is 1/3.

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At one​ store, Julia saved ​$10 with a coupon for 40​% off her total purchase. At another​ store, the ​$29 she spent on books is
barxatty [35]
Let "a" and "b" represent the values of the first and second purchases, respectively.
  0.40*(original price of "a") = $10
  (original price of "a") = $10/0.40 = $25.00 . . . . divide by 0.40 and evaluate
  a = (original price of "a") - $10 . . . . . . Julia paid the price after the discount
  a = $25.00 -10.00 = $15.00

At the other store,
  $29 = 0.58b
  $29/0.58 = b = $50 . . . . . . . divide by the coefficient of b and evaluate

Then Julia's total spending is
  a + b = $15.00 +50.00 = $65.00

Julia spent $65 in all at the two stores.
6 0
3 years ago
Simplify the expression 5^3 x 5^-5
kotykmax [81]

Answer:

\frac{1}{5^2} --- 1 over 5 squared

Step-by-step explanation:

When multiplying terms with a common base, you just add the exponents:

x^a\times x^b=x^{a+ b}

That's true even when you don't have any exponents.

5\times5=5^1\times5^1=5^{1+1}=5^2=25

\rightarrow5^3\times5^{-5}\\\rightarrow5^{3-5}\\\rightarrow5^{-2}

A negative exponent isn't fully simplified, so there's another rule to use:

x^{-y}=\frac{1}{x^y}

That is '1 over x to the y' if it's too small to read.

\rightarrow5^{-2}=\frac{1}{5^2}

4 0
2 years ago
Read 2 more answers
24 students in a class took an algebra test.
Vlada [557]

Answer:

75 % passed

25 % did NOT pass

Step-by-step explanation:

8 0
2 years ago
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A. Only Graph A is a function.
vovikov84 [41]

Answer:

b: only graph b is a function

Step-by-step explanation:

if you look closely graph a is in the middle of the boxes and all over the place, so the answer has to be b. Your welcome (pls give brainliest)

6 0
3 years ago
What prompted (caused) Gandhi to begin a campaign of passive resistance in South Africa?
Firlakuza [10]

Answer:

Asiatic Registration Bill of 1906.

Step-by-step explanation:

The first passive resistance campaign was started in Johannesburg in 1907 with, and for, the wealthy South African Indian merchants whom he had so long represented.’ Gandhi’s first passive resistance campaign began as a protest against the Asiatic Registration Bill of 1906.

5 0
2 years ago
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