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3 years ago

If there are 100 students in a class and 10% love math and the rest love reading.How many in percentage like reading.

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

90% because you just subtract 100%-90%

suter [353]3 years ago

4 0

The answer is 90%. Because 10% love math you subtract 100

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Round off 6.342 correct to 2 decimal place

maks197457 [2]

Answer:

6.34

Step-by-step explanation:

when rounding 5 or more round up 4 or less round down. hope it helps.

8 0

3 years ago

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The length of a parasite in experiment A is 2.5 × 10–3 inch. The length of a parasite in experiment B is 1.25 × 10–4 inch. How m

torisob [31]

We have been given that :-

The length of a parasite in experiment A is $2.5 \times 10^{-3}$

The length of a parasite in experiment B is $1.25 \times 10^{-4}$

Let us write the the length of the parasite in experiment A in the exponent of -3.

$1.25 \times 10^{-4}= 0.125 \times 10^{-3}$

Clearly, the length of parasite in experiment A is greater than the length of parasite in experiment B.

The difference in the length is given by

$2.5 \times 10^{-3} -0.125 \times 10^{-4}$

$=2.375 \times 10^{-3}$

Therefore, the length of the parasite in experiment A is $=2.375 \times 10^{-3}$ inches greater than the length of the parasite in experiment B.

3 0

3 years ago

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Divide 30 in the ratio of 2 : 4

cricket20 [7]

60 because 2:4 is 1/2, and when you divide by a fraction, you multiply by the reciprocal (2/1 instead of 1/2) which would be 60 in this case.

8 0

3 years ago

The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?

svet-max [94.6K]

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

$V(t) = V(0)(1-r)^{t}$

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that $r = 0.16$

$V(t) = V(0)(1-r)^{t}$

$V(t) = V(0)(1-0.16)^{t}$

$V(t) = V(0)(0.84)^{t}$

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

$V(t) = V(0)(0.84)^{t}$

$0.5V(0) = V(0)(0.84)^{t}$

$(0.84)^{t} = 0.5$

$\log{(0.84)^{t}} = \log{0.5}$

$t\log{0.84} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.84}}$

$t = 3.98$

The half life of the car is 3.98 years.

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3 years ago

Can someone help me figure out what is 5/12 of 1 ⅓ =

lorasvet [3.4K]

Answer:

Gotta multiply them

Step-by-step explanation:

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2 years ago

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