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LenKa [72]
4 years ago
12

George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrive

s just as school begins. Today he was distracted by the pleasant weather and walked the first 12 mile at a speed of only 2 miles per hour. At how many miles per hour must George run the last 12 mile in order to arrive just as school begins today?
Mathematics
1 answer:
mario62 [17]4 years ago
5 0
Note that on a normal day, it takes him <span>1/3</span> hour to get to school. However, today it took <span><span><span>1/2 mile </span><span>2 mph</span></span>=1/4</span> hour to walk the first <span>1/2</span> mile. That means that he has <span>1/3−1/4=1/12</span> hours left to get to school, and <span>1/2</span> mile left to go. Therefore, his speed must be <span><span><span>1/2 mile</span><span>1/12 hour</span></span>=<span>6 mph</span></span>, so 6 is the answer.
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Marshall earns a salary of $36,000, and each
Anton [14]

Answer:

Let Marshall's salary be  

S

m

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n

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=

$

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+

4000

n

S

j

=

$

51000

+

1500

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Set  

S

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=

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j

For convenience lets drop the $ symbol

⇒

36000

+

4000

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=

 

51000

+

1500

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1500

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and

36000

from both sides

4000

n

−

1500

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=

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=

150

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=

6

Step-by-step explanation:

7 0
3 years ago
The length of human pregnancies from conception to birth varies according to an approximately normal distribution with a mean of
g100num [7]

Answer:

68% of pregnancies last between 250 and 282 days

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 266

Standard deviation = 16

What percentage of pregnancies last between 250 and 282 days?

250 = 266 - 16

250 is one standard deviation below the mean

282 = 266 + 16

282 is one standard deviation above the mean

By the Empirical Rule, 68% of pregnancies last between 250 and 282 days

4 0
3 years ago
Solve for x and solve for y
Amanda [17]

I just noticed it is A

5 0
3 years ago
Find the output, h, when the input, x, is -18.<br> h = 17+<br> =<br> I<br> 6
Sloan [31]

The value of output, h, when the input, x, is -18 for the given equation is 14

<h3>Solving an equation </h3>

From the given question, we are to determine the value of the output, h, when the input x is -18

The given equation written properly is

h = 17 + x/6

Now, to determine the value of output, h, we will put in the value of x into the equation

h= 17 + x/6

The given value of x is -18

Thus, putting the given value of x into the equation, we get

h= 17 + x/6

h= 17 + -18/6

h = 17 + -3

h = 17 - 3

NOTE: + × - = -

h = 14

∴ h is 14 when x is -18

Hence, the value of output, h, when the input, x, is -18 for the given equation is 14

Learn more on Solving an equation here: brainly.com/question/20725347

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5 0
2 years ago
Find the area and perimeter of ABC at right. Give approximate (decimal) answers, not exact answers
Sever21 [200]

Answer:

Area of Δ ABC = 21.86 units square

Perimeter of Δ ABC = 24.59 units

Step-by-step explanation:

Given:

In Δ ABC

∠A=45°

∠C=30°

Height of triangle = 4 units.

To find area and perimeter of triangle we need to find the sides of the triangle.

Naming the end point of altitude as 'D'

Given BD\perp AC

For Δ ABD

Since its a right triangle with one angle 45°, it means it is a special 45-45-90 triangle.

The sides of 45-45-90 triangle is given as:

Leg1 =x

Leg2 =x

Hypotenuse =x\sqrt2

where x is any positive number

We are given BD(Leg 1)=4

∴ AD(Leg2)=4

∴ AB (hypotenuse) =4\sqrt2=5.66  

For Δ CBD

Since its a right triangle with one angle 30°, it means it is a special 30-60-90 triangle.

The sides of 30-60-90 triangle is given as:

Leg1(side opposite 30° angle) =x

Leg2(side opposite 60° angle) =x\sqrt3

Hypotenuse =2x

where x is any positive number

We are given BD(Leg 1)=4

∴ CD(Leg2) =4\sqrt3=6.93

∴ BC (hypotenuse) =2\times 4=8  

Length of side AC is given as sum of segments AD and CD

AC=AD+CD=4+6.93=10.93

Perimeter of Δ ABC= Sum of sides of triangle

⇒ AB+BC+AC

⇒ 5.66+8+10.93

⇒ 24.59 units

Area of Δ ABC = \frac{1}{2}\times base\times height

⇒  \frac{1}{2}\times 10.93\times 4

⇒ 21.86 units square

5 0
3 years ago
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