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sineoko [7]
3 years ago
6

Because weightlessness in outer space causes astronauts to lose bone mass, at the end of each month, an astronaut has two percen

t less bone mass than at the beginning of that month. If Lori has a 6-month mission on the International Space Station, what percentage of her bone mass will remain when she returns to Earth
Mathematics
1 answer:
vlabodo [156]3 years ago
7 0

Answer:

There will be 88.6% of her bone mass remaining when she returns to earth.

Step-by-step explanation:

According to the problem, the astronaut will have 2% less bone mass than at the beginning of that month. So let's say B_{0} is her original bone mass. In that case, her bone mass after the first month will be:

B_{1}=B_{0}(0.98)

on the second month, her bone mass will be:

B_{2}=B_{1}(0.98)=B_{0}(0.98)(0.98)=B_{0}(0.98)^{2}

on the third month, her bone mass will be:

B_{3}=B_{2}(0.98)=B_{0}(0.98)(0.98)(098)=B_{0}(0.98)^{3}

and so on, we can see a pattern here. The formula for her remaining bone mass can be generally written like this:

B_{n}=B_{0}(0.98)^{n}

where n is the number of months, so after 6 months, her remaining bone mass will be:

B_{6}=B_{0}(0.98)^{6}=0.886B_{0}

that 0.886 gives us the percentage as a decimal number. When turned into a percentage we get that:

There will be 88.6% of her bone mass remaining after 6 months.

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. A board game has dice shaped like octahedrons. Each edge of the dice has length 1 cm and the h= √ 2/ 2 cm. What is the volume
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5 0
3 years ago
If a geometric sequence has a2 = 495 and a6 = 311, approximate the value of the common ratio ???? to four decimal
shtirl [24]

Answer:

r = 0.8903

Step-by-step explanation:

given,

a₂ = 495

a₆ = 311

geometric  sequence formula

                       a_n = ar^{n-1}

                       a_2 = ar^{2-1}

                       a_2 = ar...................(1)

                       a_6 = ar^{6-1}

                       a_6 = ar^5................(2)

dividing equation (2) by (1)

                     \dfrac{a_6}{a_2} = \dfrac{ar^5}{ar}

                     \dfrac{311}{495} = r^4

                     r^4 = 0.628

                     r = 0.8903

hence, the common ratio of the geometric sequence is r = 0.8903

7 0
4 years ago
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