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3 years ago

Simplify the expression. 55−4⋅5+12

2 answers:

5 0

First we add four point five plus twelve and we get sixteen point five, and so now we subtract fifty-five minus sixteen point five and we get our answer.

Answer: 62.5

Lunna [17]3 years ago

4 0

Your answer is 62.5. Hope i helped

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If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?2

aivan3 [116]

The second equation because they all have a common factor of 3. You can get x by itself. However, with the first one, you'll end up with fractions instead.

8 0

3 years ago

Stella has a cylindrical pool. She needs 20 cubic meters of water to fill the pool to the top. The pool has a diameter of 3 m. U

Leno4ka [110]

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4 0

3 years ago

The ages of three friends are represented by three consecutive odd numbers. The sum of their ages is 63. If the youngest is repr

son4ous [18]

Answer:

x+(x+2)+(x+4)=63

Step-by-step explanation:

3 0

2 years ago

Plz help I will mark you brainlist plz hellp

Ilya [14]

Answer:Probability = 1/8.

Step-by-step explanation:

A number cube has six sides. Each side has a number from 1 to 6. This means that there are 3 even numbers (2, 4, and 6) and 3 odd numbers (1, 3, and 5). Therefore:

P(number cube lands on an even number after roll) = Number of even numbers/Number of total numbers. = 3/6 = 1/2.

Since the cube has to be rolled three times, the probability will be multiplied 3 times (assuming that each roll in independent of each other). Therefore:

P(number cube lands on an even number 3 times after roll) = 1/2 * 1/2 * 1/2 = 1/8.

Therefore, the answer is 1/8!!!

6 0

3 years ago

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = <u> -1 000.00 </u>

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago