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3 years ago

What is important about the inputs of a Funtion?

1 answer:

6 0

The input is the number you feed into the expression, and the output is what you get after the look-up work or calculations are finished. The type of function determines what inputs are acceptable; the entries that are allowed and make sense for the function.

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Find the area of the triangle.

Len [333]

Answer:

123.948 km²

Step-by-step explanation:

We can use the equation: AΔ = 1/2*ab*sinθ, where a and b are the lengths of the sides next to the known angle, and θ is the angle itself.

Plugging in the numbers we know, we get: AΔ = 1/2*22*24*sin(28°)

Simplify: AΔ = 264*.4695 (0.4695 is approximately sin(28°))

Simplify: 123.948 km²

7 0

3 years ago

What is the solution set for the inequality,

creativ13 [48]

Answer:

x≤−6

Step-by-step explanation:

Let's solve your inequality step-by-step.

4(x−3)−7x≥6

Step 1: Simplify both sides of the inequality.

−3x−12≥6

Step 2: Add 12 to both sides.

−3x−12+12≥6+12

−3x≥18

Step 3: Divide both sides by -3.

−3x

−3

≥

−3

x≤−6

6 0

3 years ago

Read 2 more answers

Watch help video

Oksanka [162]

Answer:

Step-by-step explanation:

Using the distance formula, the distance is ...

d = √((x2 -x1)^2 +(y2 -y1)^2)

d = √((-2-4)^2 +(-4-4)^2) = √((-6)^2 +(-8)^2) = √(36 +64)

d = √100 = 10

The distance between the two points is 10 units.

3 0

3 years ago

Simplify: 2(xy - 7) = [?] x = -1 and y = 3

fiasKO [112]

Answer:2(xy-7)= -20

Step-by-step explanation:

Plug in -1 for x and 3 for 3, as the question says that x and y represent those values.

2(xy-7) = ?

2((-1)(3) - 7) = ?

2(-3-7)=?

2(-10)

-20

2(xy - 7) = -20

5 0

3 years ago

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand

Tpy6a [65]

Answer:

We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

We have to test the claim that the population mean is 10.5.

Let, NULL HYPOTHESIS, $H_0$ : $\mu$ = 10.5 {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, $H_a$ : $\mu \neq$ 10.5 {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 11

s = sample standard deviation = 2

$\mu$ = population mean

n = sample of values = 16

So, test statistics = $\frac{11-10.5}{\frac{2}{\sqrt{16} } }$ ~ $t_1_5$

= 1

Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the population mean is different from 10.5.

6 0

4 years ago