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3 years ago

4(2.1k-4)=6(3.4k-2) what does (K) equal?

1 answer:

6 0

Hey there.

4(2.1k - 4) = 6(3.4k - 2); apply the distributive property to remove the parenthesis.
8.4k - 16 = 20.4k - 12; Get the variable to one side, so subtract 8.4k.
-16 = 12k - 12; isolate the variable by adding 12.
-4 = 12k; divide by 12 to solve for our variable.
k = -1/3

I hope this helps!

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

How to answer this ?

hichkok12 [17]

Answer:

a)3,6,9,12,15,18,24 b)7,14,21,28,35,52,49,56

a)16 b)40 c)24 d)36

Step-by-step explanation:

5 0

3 years ago

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Factor the question: 4x+8y+16

kirill [66]

Y= - X/2 = 2 this is the answer

8 0

3 years ago

Dewan’s bank account balance is -$16.75. He deposits checks totaling $23.59. What is his new balance?

Brilliant_brown [7]

Answer:

b because 23.59+ (-16.75) = 6.84

Step-by-step explanation:

7 0

3 years ago

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The Genetics & IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceivi

Reika [66]

Answer: The test statistic needed to test this claim= 10.92

Step-by-step explanation:

We know that the probability of giving birth to a boy : p= 0.5

i..e The population proportion of giving birth to a boy = 0.5

As per given , we have

Null hypothesis : $H_0: p\leq0.5$

Alternative hypothesis : $H_a: p>0.5$

Since $H_a$ is right-tailed , so the hypothesis test is a right-tailed z-test.

Also, it is given that , the sample size : n= 291

Sample proportion: $\hat{p}=\dfrac{239}{291}\approx0.82$

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$ , where n is sample size , $\hat{p}$ is sample proportion and p is the population proportion.

$\Rightarrow\ z=\dfrac{0.82-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{291}}}\approx10.92$

i.e. the test statistic needed to test this claim= 10.92

Critical value ( one-tailed) for 0.01 significance level = $z_{0.01}=2.326$

Decision : Since Test statistic value (10.92)> Critical value (2.326), so we reject the null hypothesis .

[When test statistic value is greater than the critical value , then we reject the null hypothesis.]

Thus , we concluded that we have enough evidence at 0.01 significance level to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.

7 0

3 years ago