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3 years ago

8

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1 answer:

alexandr1967 [171]3 years ago

4 0

Answer:

okay

Step-by-step explanation:

You might be interested in

The question is in the pic please be specific

vichka [17]

Box 1) (LxW) 20x6=120
box 2) (LxW) 15x4=60
box 1 cost) (size of box x price of box) 120x1.25=150
box 2 cost) (size of box x price of box) 60x1.25=75
subtract 150 and 75 to get 75

answer: the company is saving $75 by choosing to make 50 of box 2 instead of 50 of box 1

hope this makes sense comment if you need more explanation

8 0

3 years ago

What is the median of 0.8, 1.0, 2.7, 2.8, 3.1, 13.3

tresset_1 [31]

Answer:

$2.75$

Step-by-step explanation:

<u>Step 1: Determine the median</u>

$0.8, 1.0, 2.7, 2.8, 3.1, 13.3$

The median of a set is the middle of the set. In this instance, we have 6 numbers so we don't have a number directly in the set representing the median. Instead, we need to find the average or middle between the 3rd and 4th number. So we take 2.7 and 2.8, add them together to get 5.5 and then we divide by 2 to get 2.75 which is our median.

Answer: $2.75$

5 0

2 years ago

Read 2 more answers

Solve |t+4| > 10

diamong [38]

For
|a|>b
assume
a>b and a<-b

|t+4|>10
assume
t+4>10 and t+4<-10
solve each
minus 4 both sides

t>6 and t<-14

A is answer

7 0

4 years ago

(2x-y=24)4<br> 3x+4y=36 <br> Solve for x

dangina [55]

Answer:

x = 420/11

Step-by-step explanation:

4 0

3 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

3 years ago