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3 years ago

Pleaseeee help!!! I will mark you as brainlinest for correct answer!!!!

Mathematics

1 answer:

Maru [420]3 years ago

8 0

Evaluate the function at 2. That is f(2).
Evaluate the function at 0. That is f(0).

the average rate of change from 0 to 2 is (f(2) - f(0))/(2 - 0)

f(x) = 120(0.1)^x

f(2) = 120(0.1)^2 = 120 * 0.01 = 1.2

f(0) = 120(0.1)^0 = 120 * 1 = 120

average rate of change = (1.2 - 120)/(2 - 0) = -118.8/2 = -59.4

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The city planning committee is discussing making a bike path extension in its city park. They have a $5050 budget for the propos

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$4646.40 + $1936 To find the answer add these numbers

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2 years ago

Given that tan theta = -4/7, and 270° < theta < 360°, what is the exact value of sec theta

aleksley [76]

Answer:

sec Θ = $\frac{\sqrt{65} }{7}$

Step-by-step explanation:

Using the trigonometric identity

sec²Θ = tan²Θ + 1

Since 270° < Θ < 360° ← that is fourth quadrant, then

sec Θ > 0, thus

sec²Θ = (- $\frac{4}{7}$ )² + 1 = $\frac{16}{49}$ + 1 = $\frac{65}{49}$ , then

sec Θ = $\sqrt{\frac{65}{49} }$ = $\frac{\sqrt{65} }{7}$

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2 years ago

Jesse is playing a game with a number cube with faces numbered 1 through 6. What is the probability of rolling a number that is

lana [24]

Answer:

The probability of rolling a number that is even and a multiple of 3 is $\frac{1}{6}$

Step-by-step explanation:

The total possible outcomes of cube = { 1, 2, 3 , 4 , 5 , 6}

Now, let E: Event of getting an even number and a multiple of 3

So, out of all the outcomes, {6) is the ONLY possible favorable outcome.

$\textrm{P(E)} =\frac{\textrm{Number of favorable outcomes }}{\textrm{Total number of outcomes}}$

or, $P(E) = \frac{1}{6}$

Hence, the probability of rolling a number that is even and a multiple of 3 is $\frac{1}{6}$

8 0

3 years ago

What is 6 tens x 10=

Effectus [21]

6 tens is the same as 60 so....
60 x 10 = 600

5 0

2 years ago

Urgent. There are 12 coloured counters in a bag. The counters are black, white or grey A counter is chosen at random. The probab

Nastasia [14]

Answer:

1/12

Step-by-step explanation:

Needed information

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

The sum of the probabilities of all outcomes must equal 1

Solution

We are told that the probability that the counter is not black is 3/4.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is black by subtracting 3/4 from 1:

$\sf P(counter\:not\:black)=\dfrac34$

$\implies \sf P(counter\:black)=1-\dfrac34=\dfrac14$

We are told that the probability that the counter is not white is 2/3.

As the sum of the probabilities of all outcomes must equal 1, we can work out the probability that the counter is white by subtracting 2/3 from 1:

$\sf P(counter\:not\:white)=\dfrac23$

$\implies \sf P(counter\:white)=1-\dfrac23=\dfrac13$

We are told that there are black, white and grey counters in the bag. We also know that the sum of the probabilities of all outcomes must equal 1. Therefore, we can work out the probability the counter is grey by subtracting the probability the counter is black and the probability the counter is white from 1:

$\begin{aligned}\sf P(counter\:grey) & = \sf1-P(counter\:black)-P(counter\:white)\\\\ & =\sf 1-\dfrac14-\dfrac13\\\\ & = \sf \dfrac{12}{12}-\dfrac{3}{12}-\dfrac{4}{12}\\\\ & = \sf \dfrac{12-3-4}{12}\\\\ & = \sf \dfrac{5}{12}\end{aligned}$

7 0

2 years ago