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4 years ago

36% of what number is 18? SHOW YOUR WORK

2 answers:

8 0

So 36% of x is 18
percent is parts out of 100
36%=36/100=0.36
'of' can be translated as muitiply
0.36 times x=18
divide both sdies by .36
x=50
the number is 50

LUCKY_DIMON [66]4 years ago

7 0

Let 'x' represent the percent of 18.
Let '100' represent the whole 36.

Set up a proportion:
x/18 = 100/36
You cross multiply and get:
36x = 1800
Now divide both sides by 36:
36x/36 = 1800/36
you get:
50

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andreyandreev [35.5K]

The answer is D. 4x7+5=33 8x7+5=61. Hope it helps

5 0

3 years ago

Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represente

Nadusha1986 [10]

Answer:

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Step-by-step explanation:

Given - Ted and Jude are each saving money each month. After x months, the amount of money, in dollars, that Ted has saved is represented by the

equation T = 40x, and the amount of money that Jude has saved is represented by the equation J = 60x.

To find - Choose all of the statements that are true.

A. Each month, Jude saves two-thirds as much money as Ted saves.

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

C. The amount of money Jude saves each month is $20 more than the amount Ted saves each month.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

Proof -

Given that,

After x months,

Ted saved the amount of money, T(x) = 40x

Jude saved the amount of money, J(x) = 60x

Now,

In 1 month,

Ted saved money, T(1) = 40(1) = 40

Jude saved money, J(1) = 60(1) = 60

So,

In 1st month, Jude saved money 20 more than Ted saved.

Now,

In 2 month,

Ted saved money, T(2) = 40(2) = 80

Jude saved money, J(2) = 60(2) = 120

So,

In 2nd month, Jude saved money 40 more than Ted saved.

Now,

In 3 month,

Ted saved money, T(3) = 40(3) = 120

Jude saved money, J(3) = 60(3) = 180

So,

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option C is incorrect

Because

In 1st month, Jude saves is $20 more than the amount Ted saves

In 2nd month, Jude saves is $40 more than the amount Ted saves

Now,

In 1st month,

Ted saves = 40

and

$\frac{2}{3}(40)$ = 26.67

So, Jude will save = 40 + 26.67 = 66.67

But Jude saves 60

So,

Option A is incorrect.

i.e. A. Each month, Jude saves two-thirds as much money as Ted saves.

Now,

We can see that,

In 3 months, Ted will have saved the money = 120

In 2 months, Jude will have saved the money = 120

So,

Option B is correct.

i.e. B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

Also,

We can see that

In 1st month, Jude saved money 20 more than Ted saved.

In 2nd month, Jude saved money 40 more than Ted saved.

In 3rd month, Jude saved money 60 more than Ted saved.

So,

Option D is correct.

i.e. D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

∴ we get

True Statements are -

B. In 3 months, Ted will have saved the same amount that Jude saved in 2 months.

D. The total amount of money Jude has saved is always $20 more than the total amount Ted has saved.

7 0

3 years ago

Select the equation needed to solve the following word problem:

bija089 [108]

Answer:

Correct answer is D.

8 0

3 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

4 years ago

Sofia bought $38 worth of video games. She had a coupon that allowed her to save 8% on her purchases.

Alik [6]

Answer:34.96 (answer to what is her purchase price before tax)

Step-by-step explanation:38*(1-.08)

6 0

2 years ago