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3 years ago

What is 1/2 divided by 1/2?

Mathematics

1 answer:

schepotkina [342]3 years ago

7 0

Same as 5/5: 1
same as 29/29: 1

(1/2) / (1/2) = 1

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Work this out for me

bixtya [17]

Answer:

x = 129

Step-by-step explanation:

x and 51 are same- side exterior angles and supplementary, thus

x = 180 - 51 = 129

8 0

3 years ago

Read 2 more answers

Which formula gives the average rate of change of f(x) over the interval [8,12]?

Kruka [31]

I think the answer is C hope this would help you

5 0

4 years ago

Read 2 more answers

Only need question 2

Elza [17]

Answer:

C- all numbers less than 13.

Why these are incorrect:

A- If it was "less than or equal to" the sign would have to be "≤"

B- If it was 'greater than or equal to" the sign would have to be "≥"

D- If it was "greater than" the sign would have to be ">"

4 0

3 years ago

In each diagram, abc is a straight line. Find the unknown angles.

aleksandr82 [10.1K]

Answer:

The value of ∠b = 180°

Step-by-step explanation:

Given that;

ABC is a straight line

Another angle is 137°

Find:

The value of ∠a

The value of ∠b

Computation:

We know that, ABC is a straight line

So,

137 + The value of ∠a = 180

The value of ∠a = 180 - 137

The value of ∠a = 43°

The value of ∠b = 360 - 137 - The value of ∠a

The value of ∠b = 360 - 137 - 43

The value of ∠b = 180°

6 0

3 years ago

a. Among a shipment of 5,000 tires, 1,000 are slightly blemished. If one purchases 10 of these tires, what is the probability th

hodyreva [135]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

3 or less than 3 tires among the 10 tires should be blemished.

There are some possibilities.

1000 tires are blemished, (5000 - 1000) = 4000 tires are not blemished.

From the 5000 tires, 10 tires can be chosen in $^{5000}C_{10}$ ways.

Possibility 1: <u>3 tires are blemished.</u>

Total 10 tires can be chosen with 3 blemished tires in $^{1000}C_3 \times {4000}C_7$ ways.

Possibility 2: <u>2 tires are blemished.</u>

Total 10 tires can be chosen with 2 blemished tires in $^{1000}C_2 \times ^{4000}C_8$ ways.

Possibility 3: <u>1 tires are blemished.</u>

Total 10 tires can be chosen with 1 blemished tires in $^{1000}C_1 \times^{4000}C_9$ ways.

Possibility 4: <u>No tires are blemished.</u>

Total 10 tires can be chosen with no blemished tires in $^{1000}C_0 \times^{4000}C_{10}$ ways.

Hence, the required probability is $\frac{^{1000}C_1 \times^{4000}C_9 + ^{1000}C_0 \times^{4000}C_{10} + ^{1000}C_2 \times^{4000}C_8 + ^{1000}C_3 \times^{4000}C_7}{^{5000}C_{10} }$ .

6 0

3 years ago