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3 years ago

Find the value of x. Round to the nearest tenth.

Mathematics

2 answers:

Elenna [48]3 years ago

6 0

Answer:

Step-by-step explanation:

tan x = 4/3

tan x=1.333333

divide both sides by tan

x=1.33333/tan

x=tan inverse of 1.33333

x=53.130

x=53.13 to the nearest tenth

elena55 [62]3 years ago

4 0

Using tan(x) = 4/3

X = arctan(4/3)

X = 53.1 degrees

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What value of x makes the equation true ? 4(2x-4)=16

otez555 [7]

4(2x-4) = 16
4*2x - 4*4 = 16
8x - 16 = 16
8x = 16 + 16
8x = 32 / : 8
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7 0

3 years ago

Read 2 more answers

(25 points) Can someone please solve this I just need to see how its solved to understand

aleksley [76]

x = total amount of students in 8th Grade.

we know only one-thrid of the class went, so (1/3)x or x/3 went.

we also know 5 coaches went too, and that the total amount of that is 41.

$\bf \stackrel{\textit{one third of all students}}{\cfrac{1}{3}x}+\stackrel{\textit{coaches}}{5}=\stackrel{\textit{total}}{41}\implies \cfrac{x}{3}+5=41\implies \cfrac{x}{3}=41-5 \\\\\\ \cfrac{x}{3}=36\implies x=3(36)\implies x=108$

now, to verify, well, what do you get for (108/3) + 5?

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3 years ago

Home Depot sells 400 stoves every day. Home Depot sells 125 more microwaves per day than stoves. Which equation can be used to f

Nookie1986 [14]

Answer:

400

Step-by-step explanation:

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2 years ago

Quick please no links or anything just the straight up answer

neonofarm [45]

Answer:

Step-by-step explanation:

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3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

6 0

3 years ago