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Wewaii [24]
3 years ago
8

After Marshall's alarm went off, he spent 3/4 hr getting ready for school. He walked 1/6 hr to the bus stop, waited 1/12 hr, rod

e1/4 hr, and arrived at school at 8:30 A.M..What time did his alarm go off?
Mathematics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

7:15 am

Step-by-step explanation:

1. Add the times together by finding a common denominator

3/4 = 9/12

1/6 = 2/12

1/12 = 1/12

1/4 = 3/12

9/12 + 2/12 + 1/12 + 3/12 = 15/12

It took him 15/12 of an hour in total.

2. Convert to hours

15/12 = 1 3/12

3/12 = 1/4

1/4 of an hour is 15 minutes.

I took him a total of 1 hour and 15 minutes.

3. Subtract 1 hour and 15 minutes from 8:30

8:30 - 1:15 = 7:15

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1 meter of ribbon costs $3<br> how much do you pay for 2 3/4 <br> If correct I will brainest u
defon

Answer:

You pay $8 \frac{1}{4} for 2 \frac{3}{4} meters of ribbon

Step-by-step explanation:

Let us solve the question

∵ 1 meter of ribbon costs $3

∵ We need to find the cost of 2 \frac{3}{4} meters

→ Separate the length of the ribbon to a whole number and a fraction

∵ The length of the ribbon = 2 m + \frac{3}{4} m

→ Multiply each part by the cost of 1 meter

∵ The cost of 1 meter = 3 dollars

∴ The cost of the ribbon = 3 × 2 + 3 × \frac{3}{4}

∴ The cost of the ribbon = 6 + \frac{9}{4} dollars

→ Change the improper fraction \frac{9}{4} to a mixed number

∵ 9 ÷ 4 = 2 \frac{1}{4}

∴ The cost of the ribbon = 6 + 2 + \frac{1}{4}

∴ The cost of the ribbon = 8 + \frac{1}{4}

∴ The cost of the ribbon = 8 \frac{1}{4} dollars

∴ You pay $8 \frac{1}{4} for 2 \frac{3}{4} meters of ribbon

6 0
3 years ago
A 5 ft. tall person cast a shadow 8 ft. in length. The person is standing 32
konstantin123 [22]

Answer:

This question has insufficient information.  Assuming that the person is standing at the tip of the tree's shadow, then the tree would be  8/5 * 32 feet, or 256/5 feet, or 51.2 feet.

5 0
3 years ago
What is the slope of the line through (-3, 3) and (-1,-1)?
stira [4]
It’d be B, -2

You move right one space and then down two spaces
5 0
3 years ago
May be used to model radioactive decay. Q represents the quantity remaining after t years; k is the decay constant for plutonium
MArishka [77]

Answer:

<em>The half life of the sample is </em><em>6301 years.</em>

Step-by-step explanation:

The function used to model radioactive decay or exponential decay is,

Q(t)=Q_0e^{-k\cdot t}

Where,

Q(t) = Quantity after t time

Q₀ = Initial

k = decay constant

t = time period

As after an half life, the amount becomes half so,

\Rightarrow \dfrac{Q_0}{2}=Q_0e^{-0.00011\cdot t}

\Rightarrow \dfrac{1}{2}=e^{-0.00011\cdot t}

Taking natural log of both sides,

\Rightarrow \ln \dfrac{1}{2}=\ln e^{-0.00011\cdot t}

\Rightarrow \ln \dfrac{1}{2}={-0.00011\cdot t}\times \ln e

\Rightarrow \ln \dfrac{1}{2}={-0.00011\cdot t}\times 1

\Rightarrow {-0.00011\cdot t}=\ln \dfrac{1}{2}

\Rightarrow t=\dfrac{\ln \dfrac{1}{2}}{-0.00011}

\Rightarrow t=6301.3\approx 6301\ years

8 0
4 years ago
Simplify the expression
Katen [24]

The simplified form of the given expression would be \frac{5x^7}{2y^5}.

Hope this helps!

6 0
2 years ago
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