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antiseptic1488 [7]

2 years ago

6

Lizzie has already written 120 words of an essay. She writes 5 words per minute. Write an expression to represent the total numb

er of words she has written after her writing session today. Let m represent the number of minutes passed during today’s session.
Question

1 answer:

worty [1.4K]2 years ago

6 0

Answer:

m=24

Explanation: 120/5=24 just caculate it.

You might be interested in

Which expressions are equivalent to K/2

Fiesta28 [93]

Answer:

C and D

Step-by-step explanation:

D is the exact same.

For C this is already proven but you can test this by choosing any number for K (4) and multiply it by 1/2. You get 2 which is the same as 4/2.

3 0

2 years ago

Read 2 more answers

Determine and state the vertex of f(x) = x2– 2x – 8 using the method of completing the square.

timurjin [86]

Answer:x²-2x-8

X²-2x=8

X²= -2x + (-1)² = 8 + (-1)²

X²= -2x + 1 = 8 + 1

X²= -2x + 1 = 9

(X-1)² = 9

√(x-1)² = √9

X-1 = ± 3

X= 1 + 3= 4

X= 1 - 3= -2

X= 4 or X= -2

Step-by-step explanation:

3 0

2 years ago

Erin built a wooden box to hold hay on her farm. Hay costs $14dollar sign, 14 per cubic meter. How much will it cost to complete

sergiy2304 [10]

Here is the complete question.

Erin built a wooden box to hold hay on her farm.The box is 3 \text { m}3 m3, start text, space, m, end text long, 1 \text { m}1 m1, start text, space, m, end text wide, and 1 \text { m}1 m1, start text, space, m, end text high.Hay costs $14dollar sign, 14 per cubic meter. How much will it cost to completely fill the box with hay?

Answer:

it cost $42 to completely fill the box with hay

Step-by-step explanation:

Given that :

The length of the wooden box = 3m ; the width of the box = 1m and the height of the box = 1 m; we can easily determine the volume of the rectangular wooden box;

We all know it would assume the shape of a rectangular prism and it's volume = L × W × H

Thus; the volume of the box = . 3 m × 1 m × 1 m = 3 m³

However, the question proceeds by stating it that the wooden box cost $14 per cubic meter;

So , if 1 cubic meter = $14;

three cubic meters = $14 × 3 = $42

Hence; it cost $42 to completely fill the box with hay

8 0

2 years ago

Margo deposited $100 into a savings account earning 4.5% simple annual interest. At the end of each year, she adds $100 to her a

nadezda [96]

Answer:

principal interest total money in account

at the end of year

Year one $100 $4.5 $104.5

year two $200 $ 9 +$4.5 $ 213.5

year three $300 $ 9 +$4.5 + $13.5 $ 327

Step-by-step explanation:

Simple interest for any principal is given by

I = p* r* t/100

I = interest rate accrued on principle amount

p is the amount deposited

r is the rate of interest

t is the time period of saving

_______________________________________________

For year one

p = $100

r = 4.5%

t=1

I = 100*4.5*1/100 = 4.5

_______________________________________________

For year two $100 more is added to already existing $100 in account.

p = 100 +100 = $200

r = 4.5%

t=1

I = 200*4.5*1/100 = 9

_______________________________________________

For year two $100 more is added to already existing $200 in account after two years.

p = 100 +100 +100 = $300

r = 4.5%

t=1

I = 300*4.5*1/100 = 13.5

_______________________________________________

There fore total money in Margo account is

$300 saving deposited by her

$4.5 + $9 + $13.5 = $27 (interest accrued in three time)

Formulating the results in tabular form

principal interest total money in account

at the end of year

Year one 100 4.5 104.5

year two 200 9 +4.5 213.5

year three 300 9 +4.5 + 13.5 327

7 0

2 years ago

Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b

horrorfan [7]

Answer:

a. $X\sim N(\mu = 6.1, \sigma = 1.9)$ b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is $z_{1} = (6.4-6.1)/1.9 = 0.1579$ and the z-score related to 7 kg is $z_{2} = (7-6.1)/1.9 = 0.4737$ , we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0

2 years ago