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Ostrovityanka [42]

3 years ago

15

In a school 30% of year 10 students are boys. Of these 25% chose to take PE GCSE. Given there are 240 students in year 10 how ma

ny boys chose PE GCSE?

1 answer:

Arada [10]3 years ago

6 0

So there are 240 students in the class

And if 30% are boys that means

$\frac{30}{100} \times 240 = 72$

so out of the 72 boys if 25% take it

$72 \times \frac{25}{100} = \frac{72}{4} = 18$

Ans: 18 boys take PE GCSE

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Cat este 2^10????plass

lina2011 [118]

2 ^ 10 = 2¹⁰ = 1,024 .

2 ^ 10 este (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 ) (10 veces) tambien.

3 0

3 years ago

A spinner has 16 equal sides section 12 other side sections are purple what is the probability ti The Spinners will land on purp

pickupchik [31]

Answer:

12/16 or 3/4 or 75%

Step-by-step explanation:

4 0

3 years ago

13s - r^2; if r = 9 and s = 6

FromTheMoon [43]

Answer:

27

Step-by-step explanation:

r = 9

s = 6

r^2

= r x r

= 9 x 9

= 81

13s

= 13 x 6

= 108

13s - r^2

= 108 - 81

= 27

8 0

3 years ago

Read 2 more answers

Lines k and n intersect on the y-axis

avanturin [10]

a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in <u>slope-intercept formula</u>, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of <u>k and m.</u>

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

A similar problem is given at brainly.com/question/16302622

7 0

2 years ago

Find the distance between points M (6,16) and Z (-1,14) to the nearest tenth, please help step by step

insens350 [35]

The distance between 2 points P(a, b) and Q(c,d) is given by the formula:

$\displaystyle{d= \sqrt{(b-d)^2+(a-c)^2}$ ,

Apply the formula for the points M(6, 16) and Z(-1, 14):

$\displaystyle{d= \sqrt{(16-14)^2+(6-(-1))^2}= \sqrt{2^2+7^2}= \sqrt{4+49}= \sqrt{53}\approx 7.28$

which rounded to the nearest tenth is 7.3 (units)

Answer: 7.3 units

3 0

3 years ago