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sergey [27]
1 year ago
5

Multiply. [−12⋅47]⋅[−14⋅(−0.1)] What is the product?

Mathematics
1 answer:
yan [13]1 year ago
8 0

Answer:

-789.6

Step-by-step explanation:

I don't know if this is right

You might be interested in
A parcel has a mass of 500g. Calculate its weight. (Assume the gravitational field strength is 10N/kg).
Elan Coil [88]

Answer:

The weight of the parcel is 5 N

Step-by-step explanation:

we know that

Weight can be calculated using the equation:

 W=mg

where

W ----> weight (W) is measured in newtons (N)

m ----> mass (m) is measured in kilograms (kg)

g ----> gravitational field strength (g) is measured in newtons per kilogram (N/kg)

In this problem we have

g=10\ N/kg

m=500\ g

Remember that

1\ kg=1,000\ g

To convert g to kg

Divide by 1,000

so

m=500\ g=500/1,000=0.5\ kg

substitute the values in the equation

W=(0.5)(10)=5\ N

therefore

The weight of the parcel is 5 N

5 0
2 years ago
charles deposited $12,000 in the bank. He withdrew $5,000 from his account after one year. If he recives a total amount of $9,34
Schach [20]

Answer:

The rate of simple interest is 9%

Step-by-step explanation:

* Lets talk about the simple interest

- The simple Interest Equation (Principal + Interest)  is:

  A = P(1 + rt)  , Where

# A = Total amount (principal + interest)

# P = Principal amount

# I = Interest amount

# r = Rate of Interest per year in decimal r = R/100

# R = Rate of Interest per year as a percent R = r * 100

# t = Time period involved in months or years

- The rule of the simple interest is I = Prt

* lets solve the problem

- Charles deposited $12,000

∴ P = $12,000

- He withdrew $5,000 from his account after one year

- He receives a total amount of $9,340 after 3 years

∴ A = $9340 and t = 3

- Lets find the inetrest after 1 year

∵ I = Prt

∵ P = 12000

∵ t = 1

∴ I = 12000(r)(1) = 12000r

- Lets subtract the money that he withdrew

∵ He withdrew $5000

∵ He deposit at first 12000

∴ He has after the withdrew 12000 - 5000 = 7000

- The new P for the next 2 years is 7000

- This amount will take the same rate r for another two years

- The total money is $9340

∵ I = A - P

∵ A = 9340

∵ P = 7000

∴ The amount of interest = 9340 - 7000 = 2340

- The amount of interest after 3 years is 2340

- Lets find the amount of interest in the two years

∴ I = 7000(r × 2) = 14000r

- The amount of interest after the 3 years is the sum of the interest in

  the 1st year and the other 2 years

∴ 2340 = 14000r + 12000r

∴ 2340 = 26000r ⇒ divide both sides bu 2340

∴ r = 2340 ÷ 26000 = 0.09

∵ The rate R in percentage = r × 100

∴ R = 0.09 × 100 = 9%

∴ The rate of simple interest is 9%

8 0
2 years ago
The distance between B and C
alina1380 [7]

Answer:

The answer is

<h2>\sqrt{13}  \:  \: or \:  \: 3.6055 \:  \: units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{( {x1 - x2})^{2} +  ({y1 - y2})^{2}  }  \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

B ( 3 , 1 ) and C( 6 , 3)

The distance between them is

|BC|  =  \sqrt{( {3 - 6})^{2}  +  ({1 - 3})^{2} }  \\  =  \sqrt{ ({ - 3})^{2}  + ( { - 2})^{2} }  \\  =  \sqrt{9 + 4}

We have the final answer as

\sqrt{13}  \:  \: or \:  \: 3.6055 \:  \: units

Hope this helps you

5 0
2 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
2 years ago
Total employment for sheet metal workers in 2016 is projected to be 201,000. if this is a 6.3% increase from 2006, approximately
fomenos

Answer:

Option d.

Step-by-step explanation:

Total employment for sheet metal workers in 2016 is projected to be = 201,000

This is a 6.3% increase from 2006.

Let the number of workers in 2006 be n

n + (6.3% of n) = 201,000

n + ( 0.063n) = 201,000

1.063n = 201,000

n = \frac{201,000}{1.063}

n = 189,087. 4882 ≈ 189,000

the total employment in 2006 was approximately 189,000

Option d is the answer.

8 0
2 years ago
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