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3 years ago

10. Identify the pronoun in this sentence.

Mathematics

2 answers:

Over [174]3 years ago

8 0

Answer:

2:we.............

brilliants [131]3 years ago

4 0

Answer:

The answer is we. Pronouns are I, you, he/she/it, we, you, they.

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Question 10 Please Help

svet-max [94.6K]

Answer:

1)· 5x + 2y = 9. First we solve for y. 2y= 9 -5x. y=(9-5x)/2. Now that we have the value of y. We substitute on the original equation and resolve. 5x + 2y = 95x + 2y = 9 5x + 2(9-5x)/2 = 9 5x + 9 - 5x = 9 9 = 9

That would be x = 1

Now substitute and resolve to find y.

5(1) + 2y = 9

5 + 2y = 9

2y = 4

y = 2

So our answer x=1 and y = 2. (1,2)

Proof :

5(1) + 2(2) = 9

5+ 4 = 9

9 = 9

Step-by-step explanation:

hoped it helped for the first one i didnt now the second one

4 0

2 years ago

The number of ducks and pigs in a field totals 38. The total number of legs among them is 94 assuming each duck has exactly two

Reika [66]

Answer:

22 pigs and 3 ducks

Step-by-step explanation:

22 pigs would make 88 legs and 3 ducks will make 6 legs so u add that together and you get 94 legs (smile) ik i'm smart lol

8 0

3 years ago

What is the value of y in the equation 2(3y + 7 + 5) = 196 − 16? (1 point) Group of answer choices 13 14 26 28

Tema [17]

Answer:

Option c) is correct.

The value of y in the given equation is 26 ie., y=26

Step-by-step explanation:

Given equation is

$2(3y+7+5)=196-16$

Now to the find the value of y in the above equation.

To find y simplify the above equation

$2(3y+7+5)=196-16$

$2(3y+12)=180$

Now apply distributive property

$6y+24=180$

$6y=180-24$

$=\frac{156}{6}$

Therefore y=26

Option c) is correct.

The value of y in the given equation is 26 ie., y=26

3 0

3 years ago

Read 2 more answers

The diameter of cold virus cell is about 3x10-9 meter,and the diameter of a streptoccus bacterium cell is about 9x10-7 meter. Wh

Archy [21]

Here we need to compare the orders of magnitude of two measures.

We will see that the bacteria is 300 times larger than the virus.

We know that:

The diameter of the virus-cell is:

$D_v = 3 \cdot 10^{-9} m$

The diameter of the bacteria cell is:

$D_b = 9 \cdot 10^{-7} m$

So, to compare them, first notice that both are in the same units, meters, so we only need to compare the numbers.

Is common sense to identify the one with the largest exponent as the largest number, and the largest exponent is -7, thus the bacteria should be the larger one.

But let's prove this with math, remember the property:

$\frac{a^n}{a^m} = a^{n - m}$

Let's take the quotient between the diameters and see what we get, I will use the diameter of the bacteria in the numerator, thus if the quotient is larger than 1, it would mean that the bacteria is greater and by how much.

$quotient = \frac{ 9 \cdot 10^{-7} m}{ 3 \cdot 10^{-9} m} = \frac{9}{3} \cdot 10^{-7 + 9}\\\\= 3*10^2$

So we can say that the bacteria is 3*10^2 = 3*100 = 300 times larger than the virus.

If you want to learn more, you can read:

brainly.com/question/4953281

8 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