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2 years ago

ONLY ANSWER IF YOU KNOW HOW TO DO THIS.

Mathematics

1 answer:

erik [133]2 years ago

7 0

1. Find the SA... (30 x 20 x 2) + (20 x 4 x 2) + (30 x 4 x 2) = 1600 sq. ft.

2. 1600/300 = (6) gallons of paint

3. 30 x 20 x 4 = 2400 sq. ft.

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What values for theta (0 less than or equal to theta less than equal to 2pi) satisfy the equation?

Novosadov [1.4K]

Solve this as you would an equation that does not involve trig. Don't let the trig scare you. If you had to solve 2x+8=0, the first thing you would do is factor out the common 2. In our equation, we have a common cos theta. I'm going to use beta as my angle. When we factor out beta, here's what we have. $cos \beta (2sin \beta +1)=0$ . The Zero Product Property tells us that at least one of those factors has to equal zero. So we set them both equal to zero and solve. Let's get the equations first, then we will need our unit circle. First equation set to equal zero is $cos \beta =0$ . On our unit circle, cos is the value inside the parenthesis that is in the x position within our coordinate. Look at all those coordinates as you go around the unit circle once (once around is equivalent to 2pi). You will find that the the cos is 0 at $\frac{ \pi }{2}, \frac{3 \pi }{2}$ . The next equation is $2sin\beta +1=0$ . Move the 1 over by subtraction and divide by 2 to get $sin \beta =- \frac{1}{2}$ . Same as before, go around the unit circle one time and look to see where the coordinate in the y place is -1/2. Sin corresponds to the y coordinate. You will find that sin is -1/2 at $\frac{7 \pi }{6}, \frac{11 \pi }{6}$ . And there you go! Trig is so much fun!!!

7 0

3 years ago

Help me please. ------------------------------------------------

ladessa [460]

Answer:

Alicia

Step-by-step explanation:

First, convert the mixed fractions into improper fractions

Alicia

Distance: 5 1/2 = 11/2 = 5.5 miles

Time: 1 2/5 = 7/5 = 1.4 hours

Carlos

Distance: 8 1/8 = 65/8 = 8.125 miles

Time: 2 1/2 = 5/2 = 2.5 hours

Henry

Distance: 3 2/3 = 11/3 = 3.6 reoccuring miles

Time: 1 1/10 = 11/10 = 1.1 hours

Zoe

Distance: 4 2/3 = 14/3 = 4.6 reoccuring miles

Time: 1 3/10 = 13/10 = 1.3 hours

After that, divide the distance by the time

Alicia = 5.5 / 1.4 = 3.93 ( 3 s.f )

Carlos = 8.125 / 2.5 = 3.25

Henry = 3.6 re / 1.1 = 3.3 reoccuring

Zoe = 4.6 / 1.3 = 3.54 ( 3 s.f )

Looking at these answers, we determine which rate is the greatest

Alicia = 3.93

Carlos = 3.25

Henry = 3.3 re

Zoe = 3.54

Alicia hiked the highest rate.

5 0

3 years ago

Solve for x: 2(x-3)+5=4-7(2x-1)

Marat540 [252]

Solve for x by simplifying both sides of the equation , then isolating the variable
your answer is x=3/4

3 0

2 years ago

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are cla

AnnyKZ [126]

Answer:

(a) Name: Multinomial distribution

Parameters: $p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

(b) Range: $\{(x,y,z)| x + y + z=20\}$

Parameters: $p_1 = 5\%$ $n = 20$

$(d)\ E(x) = 1$ $Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1) = 0.7359$

$(h)\ E(Y) = 17$

Step-by-step explanation:

Given

$p_1 = 5\%$

$p_2 = 85\%$

$p_3 = 10\%$

$n = 20$

$X \to$ High Slabs

$Y \to$ Medium Slabs

$Z \to$ Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

$X \to$ Number of voids considered as high slabs

$Y \to$ Number of voids considered as medium slabs

$Z \to$ Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

$p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

And the mass function is:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

$n = 20$

The number of voids (x, y and z) cannot be negative and they must be integers; So:

$x + y + z = n$

$x + y + z = 20$

Hence, the range is:

$\{(x,y,z)| x + y + z=20\}$

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

$p_1 = 5\%$ and $n = 20$

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

$p_1 = 5\%$ and $n = 20$

$E(x) = p_1* n$

$E(x) = 5\% * 20$

$E(x) = 1$

$Var(x) = E(x) * (1 - p_1)$

$Var(x) = 1 * (1 - 5\%)$

$Var(x) = 1 * 0.95$

$Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3)$

In (b), we have: $x + y + z = 20$

However, the given values of x in this question implies that:

$x + y + z = 1 + 17 + 3$

$x + y + z = 21$

Hence:

$P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P{X \le 1, Y = 17, Z = 3)$

This question implies that:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)$

Because

$0, 1 \le 1$ --- for x

In (e), we have:

$P(X = 1, Y = 17, Z = 3) = 0$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0$

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

In (a), we have:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

So:

$P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}$

Expand

$P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}$

Using a calculator, we have:

$P(X=0; Y=17; Z = 3) = 0.07195$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

$P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1)$

This implies that:

$P(X \le 1) = P(X = 0) + P(X = 1)$

In (c), we established that X is a binomial distribution with the following parameters:

$p_1 = 5\%$ $n = 20$

Such that:

$P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}$

So:

$P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}$

$P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}$

$P(X=0) = 1 * 1 * (95\%)^{20}$

$P(X=0) = 0.3585$

$P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}$

$P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}$

$P(X=1) = 0.3774$

So:

$P(X \le 1) = P(X = 0) + P(X = 1)$

$P(X \le 1) = 0.3585 + 0.3774$

$P(X \le 1) = 0.7359$

$(h)\ E(Y)$

Y has the following parameters

$p_2 = 85\%$ and $n = 20$

$E(Y) = p_2 * n$

$E(Y) = 85\% * 20$

$E(Y) = 17$

8 0

3 years ago

In the parking lot shown, the lines

jolli1 [7]

22 is the answer because I just did it

8 0

3 years ago