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3 years ago

9

Mathematics

1 answer:

Gwar [14]3 years ago

6 0

Answer:

It is A and I am positive its A

Step-by-step explanation:

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a teacher guessed 30 students would ask for help. However, there were actually 35 students who came to get help. What was her pe

almond37 [142]

Answer:13%

Step-by-step explanation:

Lets call the Teacher Mrs.Lukens and we put the number into a calculator and subtract 35-13% you have it but as it was a guess you will never know.

3 0

3 years ago

Write the denominator of the rational number257/500 in the form 2 m x 5 n , m and n are negative integers Hence write its decima

maw [93]

Answer:

0.514

Step-by-step Explantion:

Denominator = 500 = 2^2 * 5^3

257/500 = 257/2^2*5^3 = 2*257/2*2^2*5^3 = 514/2^3*5^3 = 514/(2*5)^3 = 514/10^3 = 0.514

6 0

3 years ago

What is the unit rate if there are 258 purple polka-dot pencils in 3 purple polka-dot purses?

stiks02 [169]

Unit rate means one. So the best way to solve this is using a proportion over one. If there are 258 pencils in 3 purses 258/3 then how many are in one purses x/1. Cross multiply next. 258/3 * x/1. You get 3x and 258. To get the x alone, divide both sides by three. Your answer is 86 pencils in one purse.

4 0

3 years ago

The sum of 4 and 5 is subtracted from one-third of the difference of 40 and 4.

xxTIMURxx [149]

Answer:

Sum of 4 and 5 → 4 + 5 = 9

Difference of 40 and 4 → 40 – 4 = 36

One-third 1/3 of 36 → 12

→ 12 – 9

» 3

3 0

2 years ago

Read 2 more answers

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

4 years ago