All categories

2 years ago

I need help on this asap

Mathematics

2 answers:

Bad White [126]2 years ago

7 0

Answer:

2nd option (y = 3/5x-2)

Step-by-step explanation:

First of all, we already know the y intercept is -2, because it shows it on the graph. Also, starting from -2, you can count 5 units on the graph to the right (run), and 3 units above (rise) which will get you 3/5 due to rise over run. This is a positive slope so there is no need to put a negative behind 3/5x.

hope it helped

yaroslaw [1]2 years ago

3 0

The answer is y=3/5x-2

You might be interested in

Find the recursive rule, explicit rule, and f(20)

DiKsa [7]

Answer:

Recursive:

$f(1)=35, f(n)=f(n-1)+10$

Explicit:

$f(n)=35+10(n-1)$

And the 20th term is 225.

Step-by-step explanation:

We have the sequence:

35, 45, 55, 65.

Notice that each subsequent term is 10 more than the previous term.

Therefore, our common difference is (+)10.

Recursive Rule:

The standard format for the recursive rule is:

$f(n)=a, f(n)=f(n-1)+d$

Where a is the initial term and d is the common difference.

From our sequence, we know that a the initial term is 35.

And as determined, our common difference d is 10.

Substitute. Hence, our recursive rule is:

$f(1)=35, f(n)=f(n-1)+10$

Explicit Rule:

The standard format for the explicit rule is:

$f(n)=a+d(n-1)$

Where a is the initial term and d is the common difference. So, let’s substitute 35 for a and 10 for d. Hence, our explicit formula is:

$f(n)=35+10(n-1)$

Now, let’s find the 20th term. We will utilize the explicit rule since the recursive rule can get tedious. Substitute 20 for n because we would like to 20th term. Thus:

$f(20)=35+10(20-1)$

Evaluate:

$\begin{aligned} f(20)&=35+10(19) \\ f(20)&=35+190 \\ f(20)&=225 \end{aligned}$

Hence, the 20th term is 225.

5 0

2 years ago

you are asked to solve for y in the equation -3=5y-2x , given that x= -4 what is the first step you will take

Genrish500 [490]

First step is replacing x with -4. => -3=5y+8 => 5y= -8-3 => y =-11/5

8 0

2 years ago

keisha drove 380 miles using 14 gallond of gas. At this rate, how many gallons of gas would she need to drive 418 miles?

pychu [463]

14gallons=380miles

14/380=<span>0.03684210526gallons/mile
</span>
418*<span>0.03684210526=15.4gallons
</span>
or you could cross multiply

14gallons=380miles
x gallons=418miles

$\frac{14}{380} = \frac{x}{418}$

380x=5852
divide both sides by 380
x=15.4 gallons

8 0

2 years ago

Marshall divides his money into four categories. He saves 1/3 of his money. He gives 1/6 of his money to a charity. He uses 1/14

Len [333]

1/3 + 3/7
save + fun + extra

(7/21 + 9/21) = 16/21 on save and fun

7 0

3 years ago

Read 2 more answers

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

2 years ago