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3 years ago

There are 8 students on the minibus.five of the students are boys.what fraction of the students are boys

Mathematics

1 answer:

PolarNik [594]3 years ago

4 0

5/8 of the students are boys

5 representing boys
8 representing all students

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Which are partial products for 68 × 43?

blagie [28]

Partial products are the products obtained during the intermediate stages in order to complete a multiplication process.

Consider 68 $\times$ 43, we have to determine the partial products in this.

Now, $68 \times 43$

Expanding this, we get

$(60+8) \times (40+3)$

= $(60\times 40) +(60 \times 3) + (8 \times 40) + (8 \times 3)$

= 2400 + 180 + 320 + 24

= 2924

Hence, $8 \times 40=320$ and $8 \times 3 = 24$ are the required partial products in the product of 68 and 43.

So, Option 3 and 4 are the correct answers.

3 0

2 years ago

Write the equation of an ellipse given:

g100num [7]

The answer would be 1.23 that is the answer

7 0

3 years ago

Interest problems I’m in business Econ and I’m having help with Annuity present value and amortization if u can help me can u

Leya [2.2K]

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Step-by-step explanation:

5 0

3 years ago

Fill in the blanks please

ycow [4]

Answer:

Step-by-step explanation:

MNOP is a parallelogram Given

PM // ON opposite sides of parallelogram are parallel

∠ NOM = ∠ONP Alternate angles theorem

MN // OP opposite sides of parallelogram are parallel

∠NOP =∠ MNO Alternate angles theorem

ON = ON common to both triangles ΔOMN & ΔONP

ΔOMN ≅ ΔONP ASA congruent

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6 0

3 years ago

5. The volleyball team has a double-header on Friday. The probability that they

Nikolay [14]

Answer: 54.29%

Step-by-step explanation:

Given: The probability that they will win both games is 38%.

i.e. P( both games will win) =0.38

The probability that they will win just the first game is 70%.

P(first game will win) = 0.70

To find : P(second game will win| first game will win)

Using formula: $P(B|A)=\dfrac{P(\text{both A and B})}{P(A)}$

So, P(second game will win| first game will win) = $\dfrac{\text{ P( both games will win)}}{\text{P(first game will win)}}$

$=\frac{0.38}{0.70}\approx0.5429=54.29\%$

Hence, the required probability = 54.29%

4 0

3 years ago