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2 years ago

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Marcy determined that her father's age is the quotient of her age and 5, increased by 10. If x represents Marcy's age, write an

expression that represents her father's age?

1 answer:

Hoochie [10]2 years ago

4 0

Answer:

x / 5 + 10 or

Step-by-step explanation:

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step by step
5 times 11 - 2 times 11

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Does anyone know what graph is correct?

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The area of a regular hexagon is 14.96 cm2. What is the perimeter of the hexagon? A. 1.2 space c m B. 1.2 square root of 3 space

victus00 [196]

Answer:

14.4 cm

Step-by-step explanation:

The formula for the area of a regular hexagon in terms of its side length s is ...

... A = (3√3)/2·s²

Then the side length is ...

... s = √(2A/(3√3))

and the perimeter is ...

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of <em>k</em> is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of <em>k</em> is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

<u>Check condition 1:</u>

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

<u>Check condition 2:</u>

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of <em>y</em>.

7 0

2 years ago