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3 years ago

9

Atul, Bryony and Charlotte share an amount of money in the ratio 4:11:10

1 answer:

Alchen [17]3 years ago

8 0

4 + 11 + 10 = 25.

Assuming Bryony is the middle number, she gets 11/25 fraction of the money.

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G(x) = -0.5x^2 + 4x – 2

-BARSIC- [3]

Answer:

Step-by-step explanation:

What is the question?

g(x) = -0.5x² + 4x - 2 is a down-opening parabola. Do you need to know how to put it in vertex form?

vertex (4,6)

focus (4,5.5)

5 0

2 years ago

If l || m, classify the marked angle pair and give their relationship, then solve for X.

pickupchik [31]

Alternate exterior angles(AEA).
Given their relationship the angles are congruent.
8x-71=5x+7
3x-71=7
3x=78
X=26

6 0

3 years ago

Goryan [66]

Answer:

0

Step-by-step explanation:

$p_1:~~y = x^2+2\\p_2:~~y = 3x^2+2\\ \\ V{p_1} = \Big(-\dfrac{b}{2a}, -\dfrac{\Delta}{4a}\Big) = \Big(-\dfrac{0}{2}, -\dfrac{0^2-4\cdot 2}{4}\Big) = \Big(0,2\Big) \\ \\ Vp_2 = \Big(x_V, -\dfrac{\Delta}{4a}\Big) = \Big(0, -\dfrac{0^2-4\cdot 3 \cdot 2}{4\cdot 3}\Big) = \Big(0,2\Big) \\ \\ \\ \text{The distance is }0,~~\text{Because the vertices are equal.}$

7 0

3 years ago

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of <em>X</em> as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

4 0

3 years ago

Provide the first six terms of the following sequence:<br><br> t1=81 tn=1/3tn-1

Andrews [41]

We get tne n-th term with: tn=1/3tn-1

We know that the first term is t1=81, so 81=1/3t*1-1

81=1/3t-1

82=1/3t

82*3t=1

3t=1/82

t=1/82*3=1/246

The second term is: 1/

2*(1/246)*2)-1

We get the third term by replacing n with 3 and so on...

8 0

3 years ago

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