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3 years ago

What’s the simplified value of 12.4x(-6.3)

Mathematics

1 answer:

Eddi Din [679]3 years ago

5 0

Answer:

-78.12

Step-by-step explanation:

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A basketball player makes 90% of her free throws. what is the probability she will miss for the first time on the seventh shot?

Kobotan [32]

52.17% is the probability she will miss for the first time.

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3 years ago

A train travels 68 kilometers in 2 hours, and then 64 kilometers in 3 hours. What is its average speed

skad [1K]

Average speed is the distance traveled by an object in total and divided by the amount of time an object covers the distance. The formula is -
Average speed = total distance traveled / total time taken
= 68km+64km / 2 hrs.+3hrs.
= 132km / 5hrs.
= 26.4km/hr.
Therefore, the train's average speed is 26.4km/hr.

5 0

3 years ago

One number exceeds another by 44. the sum of the numbers is 24. what are the numbers?

baherus [9]

Let, the numbers = x,y

x+y= 24 1st eqn

x-y = 44 2nd eqn

24-y-y = 44

-2y = 44-24

y = 20/-2 = -10

substitute that in equation 1st x = 24+10 = 34

so, the numbers would be 34 & -10

4 0

3 years ago

Please help and if you dont know please don't answer

Allisa [31]

Answer:

81cm

Step-by-step explanation:

15cm + 20cm + 21cm + 25cm = 81cm

5 0

2 years ago

Read 2 more answers

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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