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3 years ago

The graph below shows the solution set of which inequality?

Mathematics

1 answer:

svp [43]3 years ago

6 0

Answer:

D x^2 smaller than or equal 4

Step-by-step explanation:

if you tried to get a number on the number line and put it to fit in the equation it will just end to equal or smaller than 4

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Triangle DEF is congruent to triangle GHI. Side DE measures 52. Side GH measures 10x-13. Side HI measures 12x-16 find the value

Lapatulllka [165]

Check the picture below.

3 0

2 years ago

PLEASE HELP URGANT The vertices of a rectangle are R(–5, –5), S(–1, –5), T(–1, 1), and U(–5, 1). A translation maps R to the poi

mote1985 [20]

Answer:

translation -(-7, -4)

5 0

3 years ago

Please answer question in picture below

Sonja [21]

9×7=63
63÷2=31.5
31.5×4=126

7×7=49

126+49=175

175 is the answer

8 0

2 years ago

{(-5,1),(-3,9), (4,6), (4, -2)} A) function B) not a function

Alexeev081 [22]

Answer:

Step-by-step explanation:

It is not a function

The 4 in the domain is used more than once

4 0

2 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: 3 tires are blemished.

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

Possibility 2: 2 tires are blemished.

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

Possibility 3: 1 tires are blemished.

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

Possibility 4: No tires are blemished.

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

2 years ago