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goblinko [34]
4 years ago
13

an object with a momentum of 12 kg m/s is travelling at 3m/s.what is its mass. Will mark as brainlest answer

Mathematics
2 answers:
raketka [301]4 years ago
7 0
Mass = 12 / 3 = 4 Kg it is
Ivenika [448]4 years ago
4 0
Mass = 12 / 3 = 4 Kg  answer
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-3(n+5)=-21 what is n​
gregori [183]

[ Answer ]

\boxed{N \ = \ 2}

[ Explanation ]

-3(n + 5) = -21

Divide Both Sides By -3

\frac{-3(n \ + \ 5)}{-3} = \frac{-21}{3}

Simplify:

N + 5 = 7

Subtract 5 From Both Sides:

N + 5 - 5 = 7 - 5

Simplify:

N = 2

\boxed{[ \ Eclipsed \ ]}

8 0
3 years ago
Read 2 more answers
Explain the steps you would take to complete this conversion problem.
gayaneshka [121]

Answer:

The answer is B

Step-by-step explanation:

What steps should be taken to complete the conversion?

Cancel the mile and kilometer units. Divide 8 by 1.61, then multiply by 1000.

Cancel the mile and kilometer units. Multiply 8 by 1.61, then multiply by 1000.

Cancel the mile and kilometer units. Multiply 8 by 1.61, then divide by 1000.

Cancel the mile and kilometer units. Multiply 8 by 1000, then divide by 1.61.

5 0
3 years ago
Read 2 more answers
PLEASE HELPPPPP
-Dominant- [34]

Answer:

126

Step-by-step explanation:

1200x.021=25.20 interest per year

25.20 x 5 years = 126

assumes no compounding

6 0
3 years ago
Evan earns $12 an hour plus $15 an hour for every hour of overtime. Overtime hours are any hours more than 30 hours for the week
Naya [18.7K]
1. M(x)=12x, with domain x < or =30, x a natural number {1,2,3,...30}

2. T(y)=12*30+15y
     
    T(y)=360+15y is the function which gives the wage after 30 hours of regular time and y of overtime.

3. T(y)=510
     360+15y=510
              15y=510-360=150
                   y=10 hours of overtime (and 30 of regular time.
6 0
3 years ago
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
4 years ago
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