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4 years ago

Given g(x)=5x−9, what is the value of g(2) ? -19 -35 2 1

Mathematics

1 answer:

Neko [114]4 years ago

5 0

Answer:

Step-by-step explanation:

g(2)=5(2)-9 Simplify

g(2)=10-9 Subtract

g(2)=1 Write value for g

g=1

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What are the missing terms? __,4,__,__,-11

Amanda [17]

9,4,-1,-6,-11
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3 years ago

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7. A wall of length 10 m was to be built across an open

Black_prince [1.1K]

Answer:

4167 bricks

Step-by-step explanation:

We first find the amount of space the wall will occupy by multiplying the dimensions given, is that;

=10m×4m×8cm

But 8cm=0.08m

This implies that,

10m×4m×8cm

=10m×4m×0.08m

$= 9.6 {m}^{3}$

The dimensions of each brick that will be used is 24 cm x12 cm x 8 cm.

We calculate the space each brick will occupy

This implies;

24 cm x12 cm x 8 cm

Converting to meters we obtain

0.24m x 0.12m x 0.08m

$=2.304 \times{10}^{ - 3}{m}^{3}$

To get the number of bricks that will be required, we divide 9.6 meters cube by

$2.304 \times {10}^{ - 3}{m}^{3}$

$\implies \frac{9.6}{2.304 \times {10}^{ - 3}} = 4166.667$

Approximately, 4167 bricks will be required.

8 0

3 years ago

Plastic toy with a mass of 18 grams occupies 9.0cm3 of space its density is

stepladder [879]

Answer:

c) 2.0 g/cm³

Step-by-step explanation:

The units of density (g/cm³) tell you to divide the mass by the volume.

... (18 g)/(9.0 cm³) = (18/9.0) g/cm² = 2.0 g/cm³

8 0

4 years ago

Let x = 4.<br><br><br><br> What is the value of y in the equation y = 15 + 11 + x?

marishachu [46]

Okay so you have to add up 15 and 11 and 4 so your answer is y=30

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3 years ago

Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5,

sweet-ann [11.9K]

Answer:

The correct option is E.

Step-by-step explanation:

It is given that Alice, Benjamin, and Carol each try independently to win a carnival game.

Let A, B and C represent the following events.

A = Alice win

B = Benjamin win

C = Carol win

$P(A)=\frac{1}{5}$ and $P(A')=1-P(A)=1-\frac{1}{5}=\frac{4}{5}$

$P(B)=\frac{3}{8}$ and $P(B')=\frac{5}{8}$

$P(C)=\frac{2}{7}$ and $P(C')=\frac{5}{7}$

We need to find the probability that exactly two of the three players will win but one will lose.

$Probability=P(A\cap B\cap C')+P(A\cap B'\cap C)+P(A'\cap B\cap C)$

$Probability=P(A)P(B)P(C')+P(A)P(B')P(C)+P(A')P(B)P(C)$

$Probability=\frac{1}{5}\cdot\frac{3}{8}\cdot\frac{5}{7}+\frac{1}{5}\cdot\frac{5}{8}\cdot\frac{2}{7}+\frac{4}{5}\cdot\frac{3}{8}\cdot\frac{2}{7}$

$Probability=\frac{3}{56}+\frac{1}{28}+\frac{3}{35}$

$Probability=\frac{7}{40}$

The probability that exactly two of the three players will win but one will lose is $\frac{7}{40}$ .

Therefore the correct option is E.

6 0

4 years ago