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leonid [27]
3 years ago
3

21% of the students had a score of 80/100 or higher on both midterms. 25% had a score of 80/100 or higher on midterm 1. what per

centage of the students who had score of 80/100 or higher on midterm 1 had a score of 80/100 or higher on midterm 2
Mathematics
1 answer:
Delvig [45]3 years ago
7 0
Let event A = high score on midterm1.
Let event B = high score on midterm2.

You want P(B|A).
  P(B|A) = P(B&A)/P(A)
  P(B|A) = 0.21/0.25
  P(B|A) = 84%

84% of students who had a high score on midterm1 had a high score on midterm 2.
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Jenna spends $15.04 at the store on notebooks and folders for school. Notebooks cost $4.85 each, and folders cost $0.89 each. Sh
cluponka [151]

Answer:

\boxed{\text{2 notebooks and 6 folders}}

Step-by-step explanation:

Let n = number of notebooks

and f = number of fodders

You have a system of two equations:

\begin{cases}(1)& 4.85n + 0.89f = 15.04\\(2) & n+ f = 8\end{cases}\\\\

\begin{array}{lrcll}(3) & f & = & 8-n &\text{Subtracted n from each side of (2)}\\& 4.85n+0.89(8-n) & = & 15.04 &\text{Substituted (3) into (1)}\\& 4.85n+7.12-0.89n & = & 15.04 &\text{Distributed 0.89}\\& 3.96n+7.12 & = & 15.04 &\text{Combined like terms} \\& 3.96n & = & 7.92 & \text{Subtracted 7.12 from each side} \\(4) & n & = & 2 & \text{Divided each side by 3.96}\\& 2+ f & = & 8 & \text{Substituted (4) into (2)}\\& f & = & 6 &\text{Subtracted 2 from each side} \\\end{array}

Jenna bought \boxed{\textbf{two notebooks and six folders} }.

Check:

\begin{array}{rclcrcl}4.85\times2+0.89\times6 & =& 15.04 & \qquad &2+6&=&8\\9.70+5.34 & = & 15.04 & \qquad &8&=&8\\15.04& = & 15.04 & & & & \\\end{array}

OK.

4 0
3 years ago
20 POINTS BRAINLIEST math
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OPTION C

Step-by-step explanation:

Given: $ y = \sqrt{36 - 72} + 5 $

This can be rewritten as: $ y = \sqrt{36(x - 2)} + 5 $

Also, since $ \sqrt{ab} = \sqrt{a}.\sqrt{b} $

$ \implies y = \sqrt{36}.\sqrt{x - 2} + 5 $

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Hence, Option C would the answer.

6 0
3 years ago
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