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nasty-shy [4]
3 years ago
8

Which equation is equal up to one-hundredth of 36? Is it 0.3*0.12 or 0.36÷0.12 or 0.3*1.2

Mathematics
2 answers:
kkurt [141]3 years ago
7 0
0.3*1.2=0.36 which is one hundredth of 36
Mumz [18]3 years ago
3 0
Its 360 beacuse u multiply all of those numbers


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A rectangular sign is 4 times its width if the perimeter is 20 inches what is the signs area.
zvonat [6]

Answer: 1,600 cubic inches

Step-by-step explanation: 20x4=80 multiply both sides 80x20=1,600 cubic inches

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Dan invests £1300 into his bank account.
alexandr402 [8]

Given parameters:

Principal  = £1300

Interest rate  = 5%

Duration  = 2yrs

Unknown:

Total amount of Dan's principal = ?

To solve this problem, we apply the equation below:

  Total investment worth  = Principal + interest

   Interest  = \frac{PRT}{100}

where P is the principal

          R is the rate

          T is the time

Now, input the parameters;

     Total investment worth  = £1300 + £ \frac{1300 x 5 x 2}{100}

                                             = £1300 + £130

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The amount after two years is £1430

7 0
3 years ago
Find the degree of the monomial.<br> -1.75K2<br> The degree is
Nikitich [7]

Answer:

2

Step-by-step explanation:

The degree of a monomial is simply the sum of all the exponents.

We can see that the only exponent belongs to k, and it is 2.

Thus, this is a second degree monomial.

The degree is 2.

And we are done!

6 0
3 years ago
Consider random samples of size 58 drawn from population A with proportion 0.77 and random samples of size 70 drawn from populat
fredd [130]

Answer:

Step-by-step explanation:

a) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as

Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}

where

p1 = sample proportion of population 1

p2 = sample proportion of population 2

n1 = number of samples in population 1,

n2 = number of samples in population 2,

From the information given

p1 = 0.77

1 - p1 = 1 - 0.77 = 0.23

n1 = 58

p2 = 0.67

1 - p2 = 1 - 0.67 = 0.33

n2 = 70

Standard error = √{(0.77 - 0.67)/[(0.77)(0.23)/58) + (0.67)(0.33)/70}

= √0.1/(0.0031 + 0.0032)

= √1/0.0063

= 12.6

the standard error of the distribution of differences in sample proportions is 12.6

b) the sample sizes are large enough for the Central Limit Theorem to apply because it is greater than 30

8 0
2 years ago
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