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Over [174]
3 years ago
5

Can the remainder in a division problem ever equal the divisor

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0
No because then somewhere in the problem you could multiply once more. like if u had 234 as ur answer, but had 2 as a remainder, u would change the 4 to a 5 (don't know if tht math is right, but the main idea of it is right)


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450 students are graduating from Big Lake High School this year. 68 percent of these students plan to go to college next year. 1
lidiya [134]
68%(plan to go to college) + 14%(plan to start working) = 82%
100% - 82% = 18% (<span>students don't know what they're going to do )
</span>
450 x 0.18 = 81

answer: 81 students <span>are not sure what they are going to do after graduating from high school</span>
5 0
3 years ago
Read 2 more answers
Please SHOW me how to complete this problem. 50 points<br><br> (10y^3-5y^2-10x+4) ÷(2y-1)
Kay [80]

Answer:

10y3 - 5y2 - 10x + 4

 ————————————————————

        2y - 1      

Step-by-step explanation:

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 

Step  2  :

Equation at the end of step  2  :

 

Step  3  :

           10y3 - 5y2 - 10x + 4

Simplify   ————————————————————

                  2y - 1      

Checking for a perfect cube :

3.1    10y3 - 5y2 - 10x + 4  is not a perfect cube

Final result :

 10y3 - 5y2 - 10x + 4

 ————————————————————

        2y - 1      

5 0
2 years ago
Read 2 more answers
Preform the indicated operation 5/8+7/12?
Vlad [161]

Answer:

5/8 + 7/12 =29/24

Step-by-step explanation:

8 0
2 years ago
What is the probability of drawing a face card from a standard deck of cards and rolling a 3 or more on a 20-sided dice?
Nastasia [14]

Answer: 20.77%

Step-by-step explanation:

There are 52 cards in a standard deck. Out of these cards, there are 12 face cards. Thus, the probability of drawing a face is 12/52.

On a 20-sided dice, there are 20 options of numbers to roll. Out of these 20 numbers, 18 of the numbers are greater than or equal to 3, making the probability 18/20, or 9/10.

To find the probability of two events, you need to multiply the probabilities together. Thus, you need to multiply 12/52 by 9/10.

(12/52) * (9/10) = 27/130

27/130 is about equal to 0.2077, or a 20.77% probability.

3 0
2 years ago
A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week
Levart [38]

Answer:

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that a simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week.

Of the 250 employed individuals​ surveyed, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                              P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of individuals who work at home at least once per week = \frac{41}{250} = 0.164

           n = sample of individuals surveyed = 250

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                             level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<em>99% confidence interval for p</em> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } , 0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } } ]

 = [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

7 0
2 years ago
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