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Luda [366]
3 years ago
13

Cara evaluated the expression below.

Mathematics
2 answers:
VashaNatasha [74]3 years ago
7 0
She multiplied 4 and 7 together and did not multiply 4 and 13

4(7 - 13) / 3 = (28 - 52)/3 <== that is what it should be....not (28 - 13) / 3
s2008m [1.1K]3 years ago
5 0

Answer:

Cara  multiplied only 4 and 7 together and did not multiply 4 and 13.

Step-by-step explanation:

Let us evaluate the given expression step by step,

\frac{4(7-13)}{3} +(-4)^{2} -2(6-2)\\

We can evaluate this expression in two ways; either by subtracting -13 from 7 and then multiply by 4 or by multiplying 4 to 7  and -13.

\frac{28-52}{3} +16-2(4)

\frac{-24}{3} +16-8

-8+16-8=-16+16=0

Now let us see where Cara made an error. We can see Cara just multiplied only 4 and 7 together and did not multiply 4 to 13.

Therefore, Cara made error in her first step.

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3 years ago
Craig solicited donations for his school's jump-rope a-thon. He collected $13.00 in fixed donations and pledges totaling $1.50 f
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What is the answer to thisss?
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The value of x is 14

<h3>How to solve for x?</h3>

From the question, the lines from the center of the circle to the chords have equal measures of 9 units

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If a function is periodic, with period c, then so is its derivative?
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Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur
stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem:

  • Event A: Three heads.
  • Event B: Fair coin.

The probability associated with 3 heads are:

  • 0.5^3 = 0.125 out of 0.5(fair coin).
  • 1 out of 0.5(biased).

Hence:

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The probability of 3 heads and the fair coin is:

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Then, the conditional probability is:

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A similar problem is given at brainly.com/question/14398287

4 0
3 years ago
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