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sergij07 [2.7K]
3 years ago
10

Can someone please help me

Mathematics
1 answer:
ss7ja [257]3 years ago
6 0

Answer:

a.) 1/2

b.) 1/8

c.) 5/8

d.) 1/2

Step-by-step explanation:

for A it is asking for probability of 1 and is 1/2 of the circle

for B it is asking for  probability of 3 and it is 1/2 of and 1/4 of the circle so 1/8

for C it is asking for probability of the odd numbers and there is tow odd numbers 1 and 3 the probability of 1 is 1/2 and the probability of 3 is 1/8 adding them up will give you 5/8

for d it asks for the probability of at least 2 so therefor 2, 3, and 4 and they are 1/8 + 1/8 + for 1/4 = 1/2

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The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter
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Answer:

(1\frac{1}{2},1) - False

(4,6) - True

(18,12) -- False

(0,0) -- True

(2\frac{1}{2},3\frac{3}{4}) -- True

Step-by-step explanation:

The points are

(1\frac{1}{2},1) , (4,6), (18,12), (0,0) and (2\frac{1}{2},3\frac{3}{4}) ---- missing from the question

Given

y = 1\frac{1}{2}x

Required

Determine if each of the points would be on y = 1\frac{1}{2}x

To do this, we simply substitute the value of x and of each point in y = 1\frac{1}{2}x.

(a) (1\frac{1}{2},1)

In this case;

x = 1\frac{1}{2} and y = 1

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 1\frac{1}{2}

y = \frac{3}{2} * \frac{3}{2}

y = \frac{9}{4}

y = 2\frac{1}{4}

<em>The point </em>(1\frac{1}{2},1)<em>  won't be on the graph because the corresponding value of y for </em>x = 1\frac{1}{2}<em> is </em>y = 2\frac{1}{4}<em></em>

(b) (4,6)

In this case;

x = 4

y = 6

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 4

y = \frac{3}{2} * 4

y = \frac{3* 4}{2}

y = \frac{12}{2}

y = 6

<em>The point </em>(4,6)<em>  would be on the graph because the corresponding value of y for </em>x = 4 is y = 6

(c) (18,12)

In this case:

x = 18;y = 12

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 18

y = \frac{3}{2} * 18

y = \frac{3* 18}{2}

y = \frac{54}{2}

y = 27

<em>The point </em>(18,12)<em>  wouldn't be on the graph because the corresponding value of y for </em>x = 18<em> is </em>y = 12<em></em>

(d) (0,0)

In this case;

x =0; y = 0

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 0

y = 0

<em>The point </em>(0,0)<em>  would be on the graph because the corresponding value of y for </em>x = 0 is y = 0

(e) (2\frac{1}{2},3\frac{3}{4})

In this case:

x = 2\frac{1}{2}; y = 3\frac{3}{4}

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 2\frac{1}{2}

y = \frac{3}{2} * \frac{5}{2}

y = \frac{15}{4}

y = 3\frac{3}{4}

<em>The point </em>(2\frac{1}{2},3\frac{3}{4}) <em>  would be on the graph because the corresponding value of y for </em>x = 2\frac{1}{2} is y = 3\frac{3}{4}

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Answer:

The answer to your question is letter A

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WM is a secant or a chord

WO is the radius

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