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3 years ago

Which expression entered into a graphing calculator will return the probability

Mathematics

1 answer:

igomit [66]3 years ago

3 0

Answer:

Step-by-step explanation:

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Plz hurry help!!!!:) What is the length of AD? Round your answer to the nearest hundredth.

jolli1 [7]

Answer:

AD=5.82 units

Step-by-step explanation:

tan 31=3/BD

0.6009=3/BD

BD=5

AD^2=3^2 + 5^2

AD^2=9+25=34

AD=5.82

8 0

3 years ago

Read 2 more answers

30km=?miles 44,000in=?cm 21ft=?m 0.75miles=?dam 133yards=mm Please show your work

kherson [118]

So there is this chart my teacher showed me to help find the answers to your questions I will send a picture but here are the answers

30km is 18.642m
44,000in is 111760cm
21ft is 6.401m
0.75m I’m not for sure sorry but
133yards is 121615.2mm

I hope that helps sorry I did not understand the one question 0.75m

5 0

3 years ago

Solve for x. 13(x-3) = 39 X=1 X=4 x=6 x=10

Valentin [98]

13 factored out is 13x - 39 add 39 to both sides it equals 13x=78 and 78 divided by 13 is 6 the anwser is x=6

7 0

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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