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3 years ago

For each object identify two shapes that can be combined to model the object

Mathematics

1 answer:

ira [324]3 years ago

3 0

A). The second one and the fourth one combined make the object ( a )

B). First one and the last one combined make the object ( b )

C). First one and the third one combined make the object ( c )

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2/3 divided into 4 groups, what fraction represents the answer to this question?

kondaur [170]

2/3 ÷ 4 = 2/3 * 1/4 = 2/12 = 1/6

5 0

3 years ago

Read 2 more answers

Can someone please help me

djyliett [7]

Slope = (10 - 5)/(-9 + 15) = 5/6

y = mx + b where m = slope and b = y-intercept

b = y - mx

b = 5 - (5/6) (-15)

b = 5 +12.5

b = 17.5

Equation y = 5/6 x + 17.5

Y-intercept = 17.5

X-intercept when y = 0

5/6 x + 17.5 = 0

5/6 x = -17.5

x = (-17.5) (6/5)

x = -21

Answer

Y intercept (0, 17.5 )

X intercept (-21 , 0)

3 0

3 years ago

A guy wire 30 ft long supports an antenna at a point that is 24 ft above the base of the antenna.

GalinKa [24]

Use Pythagorean Theorem
c2 = a2 + b2
152 = a2 + 12
a2 = 152 - 122
=225 - 144 = 81
a = 9 ft
Distance from base of antenna = 9ft

4 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$