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DIA [1.3K]
3 years ago
7

Troy runs marathons, which are about 26 miles long. He usually runs 4 marathons a year, and he has been doing this for 11 years.

Troy estimates he has run a little over 100 miles during all these marathons. Is that a good estimate?
Yes


No it is much high



No it is much low
Mathematics
1 answer:
aliya0001 [1]3 years ago
4 0
It is much too low as if he runs 4 every year for the past 11 years he’s ran about 44 marathons and each marathon is 26 miles
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3 0
3 years ago
Santo made 4 less than twice as many cupcakes as Lynn. Write an equation to represent the number of cupcakes that Santo made.
pickupchik [31]

Santo made 4 less than twice as many cupcakes as Lynn.


x * 2 - 4

5 0
3 years ago
5-2(4a+1)+3a=13 <br><br> A= 2/3<br> A= -2<br> A= -6/5<br> A=2
VMariaS [17]

Answer:

                   a = -2

Simplifying 5 + -2(4a + 1) + 3a = 13 Reorder the terms: 5 + -2(1 + 4a) + 3a = 13 5 + (1 * -2 + 4a * -2) + 3a = 13 5 + (-2 + -8a) + 3a = 13 Combine like terms: 5 + -2 = 3 3 + -8a + 3a = 13 Combine like terms: -8a + 3a = -5a 3 + -5a = 13 Solving 3 + -5a = 13 Solving for variable 'a'. Move all terms containing a to the left, all other terms to the right. Add '-3' to each side of the equation. 3 + -3 + -5a = 13 + -3 Combine like terms: 3 + -3 = 0 0 + -5a = 13 + -3 -5a = 13 + -3 Combine like terms: 13 + -3 = 10 -5a = 10 Divide each side by '-5'. a = -2

6 0
3 years ago
Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so
IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

<em><u>Solution:</u></em>

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

<em><u>Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars</u></em>

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

14 \times c + 5 \times v=119

14c + 5v = 119 ------- eqn 1

<em><u>Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars</u></em>

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

10 \times c + 10 \times v = 130

10c + 10v = 130 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

<em><u>Subtract eqn 2 from eqn 3</u></em>

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

<em><u>Substitute c = 6 in eqn 1</u></em>

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0
3 years ago
A certain genetic condition affects 8% of the population in a city of 10,000. Suppose there is a test for the condition that has
enot [183]

Solution:

Population in the city= 10,000

As genetic condition affects 8% of the population.

8 % of 10,000

=\frac{8}{100}\times 10,000=800

As, it is also given that, there is an error rate of 1% for condition (i.e., 1% false negatives and 1% false positives).

So, 1% false negatives means out of 800 tested who are found affected , means there are chances that 1% who was found affected are not affected at all.

So, 1% of 800 =\frac{1}{100}\times 800=8

Also,  1% false positives means out of 10,000 tested,[10,000-800= 9200] who are found not affected , means there are chances that 1% who was found not affected can be affected also.

So, 1% of 9200 =\frac{1}{100}\times 9200=92

1. Has condition Does not have condition totals  = 800

2. Test positive =92

3. Test negative =8

4. Total =800 +92 +8=900

5. Probability (as a percentage) that a person has the condition if he or she tests positive= As 8% are found positive among 10,000 means 9200 are not found affected.But there are chances that out of 9200 , 1% may be affected

=\frac{\text{1 percent of 9200}}{9200}\\\\ \frac{\frac{1}{100}\times 9200}{9200}=\frac{92}{9200}\\\\ =0.01

that is Probability equal to 0.01 or 1%.

7 0
3 years ago
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