In a test,

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

7 0

Riya fails in the exam by 2 marks.

Step-by-step explanation:

Step 1: Find the total marks required to pass.

Total marks in Section A = 80 marks

Total marks in Section B = 120 marks

Total marks in the test = 120 + 80 = 200 marks

Percentage required to pass = 65%

⇒ Required marks to pass = 65% of 200 = 65/100 × 200 = 130 marks

Step 2: Find marks scored in Section A and Section B.

Marks scored in Section A = 55% = 55% of 80 = 55/100 × 80 = 44 marks

Marks scored in Section B = 70% = 70% of 120 = 70/100 × 120 = 84 marks

Step 3: Find total marks scored.

Total marks scored = 44 + 84 = 128 marks

128 < 130 ⇒ Riya fails in the test.

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A group consists of five men and five women. Three people are selected to attend a conference.

nadya68 [22]

Answer:

a) 120

b) 10

c) 1/12

Step-by-step explanation:

The number of ways that a sample of r can be selected from a population of n is:

nCr = n! / (r! (n−r)!)

a) 3 people selected from a group of 10

₁₀C₃ = 120

b) 3 women selected from a group of 5

₅C₃ = 10

c) Of the 120 committees that can be chose, 10 are all women. So the probability is 10/120 = 1/12.

6 0

3 years ago

Find an equation for the line perpendicular to y=−15x+3 with x-intercept at x = 3.

Alex Ar [27]

bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of the provided line then

$\bf y=\stackrel{\stackrel{m}{\downarrow }}{-15}x+3\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-15\implies -\cfrac{15}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{1}{15}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{15}\implies \cfrac{1}{15}}}$

well, we know the x-intercept is at x = 3, recall when a graph intercepts the x-axis y = 0, so this point is (3 , 0). Then we're really looking for the equation of a line whose slope is 1/5 and runs through (3 , 0).

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{0})~\hspace{10em} slope = m\implies \cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=\cfrac{1}{5}(x-3)\implies y=\cfrac{1}{5}x-\cfrac{3}{5}$