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IRISSAK [1]
3 years ago
15

A doctor ordered a patient to exercise. The patient ealked for 147 minutes in 2 weeks. The ratio of the time for the first week

to the time for the second week is 8:13. How many minutes did the patient walk each week?
Mathematics
1 answer:
Ray Of Light [21]3 years ago
4 0
1) 8x + 13x = 147
    21x = 147
    x = 147 : 21
    x = 7 (one part)
2) 7 * 8 = 56 min. - patient walked in the first week
3) 7 * 13 = 91 min. - in the second week.
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Jenny bought a 5 notebooks and 2 pencils, and the total was 9 dollars.
weeeeeb [17]
You would solve this with simultaneous equations, so if we write it as:
5n + 2p = 9
3n + 2p = 6
(subtract)
2n = 3
÷ 2
notebooks = 1.5

Now you would substitute it in:
(3 × 1.5) + 2p = 6
4.5 + 2p = 6
- 4.5
2p = 1.5
÷ 2
pens = 0.75

So your final answer is notebooks are $1.50 and pens are $0.75, I hope this helps!
6 0
3 years ago
Esteban's credit card issuing bank charges him an APR of 16% on new purchases, and 24% on
Julli [10]

Answer: $14

Step-by-step explanation:

Hi, to answer this question we have to multiply the amount of the cash advance (700) by the APR for cash advances in decimal form (divided by 100):

700 x (24/100) = $168 (a year)

To obtain the monthly charges, we have to divide the result by the number of months in a year (12):

168 /12 = $14

Feel free to ask for more if needed or if you did not understand something.

5 0
3 years ago
If the measure of the angle in the southwest corner of the intersection of Broadway and 21st Street is approximately 105ºwhat is
Karolina [17]

Answer:

95°

Step-by-step explanation:

The angle on the 20TH street and 22ND street must add up to 180° so the answer is 95°

5 0
3 years ago
In tossing four fair dice, what is the probability of tossing, at most, one 3
11Alexandr11 [23.1K]

<u>Answer- </u>

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

<u>Solution-</u>

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48                    ( ∵ xxxx )

P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38                    ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0
3 years ago
Someone help me with this!
JulijaS [17]

Answer:

63

Step-by-step explanation:

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8 0
3 years ago
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