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3 years ago

Priya has picked 1 1/2 cups of raspberries, which is enough

Mathematics

1 answer:

MAVERICK [17]3 years ago

3 0

Can you give more information

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Evaluate

oee [108]

Answer:

-1

Step-by-step explanation:

-2+(-1)÷3

-2-1÷3

-3÷3

-1

3 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

3 years ago

What is the equation of the line that passes through (5,-2) and (-3,4)

tester [92]

(5,-2)(-3,4)
slope = (4 - (-2) / (-3 - 5) = 6/-8 = - 3/4

y = mx + b
slope(m) = -3/4
use either of ur points...(5,-2)...x = 5 and y = -2
now we sub and find b, the y int
-2 = -3/4(5) + b
-2 = -15/4 + b
-2 + 15/4 = b
-8/4 + 15/4 = b
7/4 = b

so ur equation is : y = -3/4x + 7/4 <==

3 0

4 years ago

1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following sit

Nonamiya [84]

Answer:

1 a $p -value = 0.030054$

1b $p -value = 0.0029798$

1c $p -value = 0.0039768$

2a $p-value = 0.00099966$

2b $p-value = 0.00999706$

2c $p-value = 0.0654412$

Step-by-step explanation:

Considering question a

The alternative hypothesis is H1:μ>μ0

The test statistics is z =1.88

Generally from the z-table the probability of z =1.88 for a right tailed test is

$p -value = P(Z > 1.88) = 0.030054$

Considering question b

The alternative hypothesis is H1:μ<μ0

The test statistics is z=−2.75

Generally from the z-table the probability of z=−2.75 for a left tailed test is

$p -value = P(Z < -2.75) = 0.0029798$

Considering question c

The alternative hypothesis is H1:μ≠μ0

The test statistics is z=2.88

Generally from the z-table the probability of z=2.88 for a right tailed test is

$p -value = P(Z >2.88) = 0.0019884$

Generally the p-value for the two-tailed test is

$p -value = 2 * P(Z >2.88) = 2 * 0.0019884$

=> $p -value = 0.0039768$

Considering question 2a

The alternative hypothesis is H1:μ>μ0

The sample size is n=16

The test statistic is t = 3.733

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 16 - 1$

=> $df = 15$

Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of $df = 15$ for a right tailed test is

$p-value = t_{3.733 , 15} = 0.00099966$

Considering question 2b

The alternative hypothesis is H1:μ<μ0

The degree of freedom is df=23

The test statistic is ,t= −2.500

Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is

$p-value = t_{-2.500 , 23} = 0.00999706$

Considering question 2c

The alternative hypothesis is H1:μ≠μ0

The sample size is n= 7

The test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 7 - 1$

=> $df = 6$

Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of $df = 6$ for a left tailed test is

$t_{-2.2500 , 6} = 0.03272060$

Generally the p-value for t= −2.2500 for a two tailed test is

$p-value = 2 * 0.03272060 = 0.0654412$

4 0

3 years ago

162 water bottles in 9 cases=

marishachu [46]

162/9 = 18 water bottles per case

5 0

3 years ago