SOooo .. I'm doing my Geometry DBA on edgenuity and i'm ready to quit.

The selling price of a refrigerator, is $ 796.40 . If the markup is 10 % of the dealer's cost, what is the dealer's cost of

8_murik_8 [283]

The final price (what it is selling for) is $796.40

The markup is 10% of the original price (the dealer's cost) , meaning that it is 10% more.

We need to find the original price.

We write this as an equation

The original price *110% = final price

This is because the original price is itself (100%) added with 10%

Plug in the known final price

Original Price * 110% = 796.40

Convert 110% to a decimal because the other numbers- such as the final price are also decimal numbers.

Convert 110% to a decimal by moving the decimal point up 2 spaces ( basically dividing it by 100)

110% = 1.1

So it is now

Original price *1.1 = 796.40

Divide both sides by 1.1 to isolate our unknown, the original price

Original price = $724

5 0

3 years ago

HELP PLEASE

Georgia [21]

Answer:

$\huge\boxed{\sf No}$

Step-by-step explanation:

Evan spent 20 hours doing homework last week

25 hours were spent this week.

Let's see what 125% of 20 equals:

= (125 / 100) * 20

= (125 / 10) * 2

= 125 / 5

= 25 hours

But,

It is written that Evan thought he spent 125 % more than the last week which means:

= (125% of 20) + 20

= 25 + 20

= 45 hours

He should've said that he has given 25% more time than the last week.

(Note that: 25 % of 20 equals 5 so 25 % more than last week will be equal to (25 % + 20) = (5+20) = 25)

$\rule[225]{225}{2}$

Hope this helped!

7 0

3 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) Two computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

(c) Here, also 8 computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

LET X = Number of computers classified as cutting dash edge

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0

3 years ago

Hello! I kinda need a little help with this...

Andre45 [30]

I honestly don't know or I would have replied. Sorry :(

7 0

3 years ago

Read 2 more answers

Need help

ira [324]

Formula for the volume of a cylinder is pi radius squared times height so this one would be pi times 25 times 4 which approximately equals 314

4 0

3 years ago

Read 2 more answers