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2 years ago

5

the mean mark got by 6 pupil's in a cat is 75.5 of them scored 83 73 71 67 and 69. what was the score of the sixth pupil?.what w

as the modal mark

1 answer:

vichka [17]2 years ago

8 0

Answer: score if the sixth pupil is 90

Step-by-step explanation:

Mean mark-75.5

6pupil

83+73+71+67+69+x= 75.5

363+x=75.5×6

×=453-363

×=90

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2 years ago

The quotient of (x⁴+ 5x³ 3x - 15) and (x³ - 3) is a polynomial. What is the quotient?

lesya [120]

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Step-by-step explanation:

you can use long division.

Take the leading coefficient x^4 and divide it by x^3. This results in x which is going to be the first part of you quotient. Now take that x and multiply it by the divisor (x^3 - 3). This gives you x(x^3 - 3) = x^4 - 3x. Now subtract that x^4 - 3x from the original polynomial and repeat this until you can't divide anymore

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6 0

1 year ago

are any two regular polygons similar. The area of the smaller figure is about 390 cm^2. choose the best estimate of the area of

galben [10]

Answer:

The best estimate of the area of the larger figure is $693\ cm^{2}$

Step-by-step explanation:

step 1

<em>Find the scale factor</em>

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

z----> the scale factor

x-----> the corresponding side of the larger figure

y-----> the corresponding side of the smaller figure

so

$z=\frac{x}{y}$

we have

$x=12\ cm$

$y=9\ cm$

substitute

$z=\frac{12}{9}=\frac{4}{3}$ -----> the scale factor

step 2

<em>Find the area of the larger figure</em>

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z----> the scale factor

x-----> the area of the larger figure

y-----> the area of the smaller figure

so

$z^{2}=\frac{x}{y}$

we have

$z=\frac{4}{3}$

$y=390\ cm^{2}$

substitute and solve for x

$(\frac{4}{3})^{2}=\frac{x}{390}$

$(\frac{16}{9})=\frac{x}{390}$

$x=390*16/9=693.33\ cm^{2}$

4 0

2 years ago