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3 years ago

11

Solve for u. 7/6=u+4/8 Simply the answer.

2 answers:

Katarina [22]3 years ago

7 0

Answer:

plz mark brainliest

Step-by-step explanation:

16/3 or 5 1/3

disa [49]3 years ago

4 0

Answer:

$u=\frac{16}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

<u /> $\frac{7}{6} =\frac{u+4}{8}$ <u />

<u />

<u>Step 2: Solve for </u><u><em>u</em></u>

Cross-multiply: $7(8)=6(u+4)$
Multiply: $56=6(u+4)$
Distribute 6: $56=6u+24$
Isolate <em>u</em> term: $32=6u$
Isolate <em>u</em>: $\frac{32}{6} =u$
Simplify: $\frac{16}{3} =u$
Rewrite: $u=\frac{16}{3}$

<u>Step 3: Check</u>

<em>Plug in u into the original equation to verify it's a solution.</em>

Substitute in <em>u</em>: $\frac{7}{6} =\frac{\frac{16}{3} +4}{8}$
Add: $\frac{7}{6} =\frac{\frac{28}{3}}{8}$
Divide: $\frac{7}{6} =\frac{7}{6}$

Here we see that 7/6 does indeed equal 7/6.

∴ $u=\frac{16}{3}$ is a solution of the equation.

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

(c) Name: Binomial distribution

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

Please help. I don’t understand and don’t know what do to.

Bingel [31]

9514 1404 393

Answer:

72

Step-by-step explanation:

The triangles are said to be similar. (ΔNPQ ~ ΔRSQ) That means corresponding sides have the same ratio:

NP/RS = NQ/RQ = PQ/SQ = 24/32 = 21/28 = 3/4

This ratio, or scale factor, also applies to the perimeters of the two triangles.

perimeter NPQ / perimeter RSQ = 3/4

Using the given expressions for the perimeters, we have ...

(7x +2)/(10x -4) = 3/4

We can solve this equation in the usual way to find the value of x. Then we can use that value to find the perimeter of ΔNPQ.

4(7x +2) = 3(10x -4) . . . . . multiply both sides by 4(10x -4)

28x +8 = 30x -12 . . . . . eliminate parentheses

20 = 2x . . . . . . . . . . . add 12-28x to both sides

10 = x . . . . . . . . . . . divide both sides by 10

The perimeter of ΔNPQ is ...

7x +2 = 7(10) +2 = 72

The perimeter of triangle NPQ is 72 units.

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3 years ago

Solve: 2 = (-1/5)x <br><br><br> i dont know this:(

mixer [17]

Answer: x = -10

Step-by-step explanation:

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3 years ago

Determine which consecutive integers do not have a real zero of f(x) = x3 + 5x2 - x - 6 between them.

Sunny_sXe [5.5K]

The answer is c. I took the test

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3 years ago

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melisa1 [442]

Part a: There is an 11% decrease in the price

Part b: There is an 8% decrease in the price

Part c: There is an 8% increase in the price

Part d: There is an 11% increase in the price

Part a:

Original price = $63

New price = $56.07

Change in price = 56.07-63 = - 6.93

Percentage change in price = Change in price/Original price*100

= -6.93/63*100

= - 0.11*100 = 11% decrease in price

Part b:

Original price = $37.50

New price = $34.50

Change in price = 34.5-37.5 = - 3

Percentage change in price = Change in price/Original price*100

= -3/37.5*100

= - 0.08*100 = 8% decrease in price

Part c:

Original price = $12.25

New price = $13.23

Change in price = 13.23-12.25 = 0.98

Percentage change in price = Change in price/Original price*100

= 0.98/12.25*100

= 0.08*100 = 8% increase in price

Part d:

Original price = $98

New price = $108.78

Change in price = 108.78-98 = 10.78

Percentage change in price = Change in price/Original price*100

= 10.78/98*100

= 0.11*100 = 11% increase in price

Learn more about percentage change:

brainly.com/question/14700520

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5 0

1 year ago