All categories

2 years ago

9-3 divided by 1/3 +1

Mathematics

2 answers:

polet [3.4K]2 years ago

4 0

= 9 - 3 ÷ ⅓ + 1
= 9 - 3 × 3 + 1
= 9 - 9 + 1
= 1

Andrei [34K]2 years ago

3 0

Nine thirds? If so its would be 9/3 / 1/3?

The answer would be 1/3 / 1/3+1 making it 2/3

Im not sure so dont take my word

You might be interested in

Robert is a salesperson. He receives 7% commission on each piece of electronic equipment he sells. How much commission will he r

Shalnov [3]

Answer:

He receives $476 in comission.

Step-by-step explanation:

This question can be solved using a rule of three.

$6,800 is 100% = 1. x, which is his commision, is 7% = 0.07. Then

6800 - 1

x - 0.07

x = 6800*0.07

x = 476

He receives $476 in comission.

7 0

2 years ago

A professor wants to award prizes for 1st, 2nd, 3rd, 4th and 5th in the class of 30. In how many ways can the prizes be awarded

aniked [119]

Answer:

$Selection = 17100720\ ways$

Step-by-step explanation:

Given

$Students = 30$

$Positions = 5$ i.e. first to fifth

Required

Determine number of selection

1st position = Any of the 30 students

2nd position = Any of the 29 students

3rd position = Any of the 28 students

4th position = Any of the 27 students

5th position = Any of the 26 students

Number of selection is then calculated as:

$Selection = 30 * 29 * 28 * 27 * 26$

$Selection = 17100720\ ways$

8 0

2 years ago

14.4% of a voters in middleville voted for rocky rufus for mayor. if 2500 people voted, how many votes did rocky get?

il63 [147K]

Answer:

360 votes

Step-by-step explanation:

Move the decimal to the left 2 times and you'll get .144. Multiply that by 2500, and you have your answer.

3 0

2 years ago

A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once p

____ [38]

Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

step 1:-

Given sample size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

Population proportion of employed individuals who work at home at least once per week P = $\frac{x}{n} =\frac{41}{200} =0.205$

Q=1-P= 1-0.205 = 0.705

step 2:-

Now $\sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }$

=0.0015

step 3:-

Confidence intervals

using formula

$(P - Z_∝} \sqrt{\frac{P Q}{n},} (P + Z_∝} \sqrt{\frac{P Q}{n},$

$(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887$

=0.20113,0.20887[/tex]

conclusion:-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

4 0

2 years ago

Someone help me ASAP!!!! I need this ASAP!!!!

qwelly [4]

So to solve it the equation is a squared plus b squared = c squared. So you square a and b. Then do the square root of that and should get c. C is always the largest side. C is opposite from a 90 degree angle . Hope this helps you answer your question.

5 0

2 years ago