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1 year ago

The following are the ages (years) of 5 people in a room:

Mathematics

1 answer:

strojnjashka [21]1 year ago

7 0

Answer:

67 years old. (you only gave 4 people and missed out one person)

Step-by-step explanation:

take all the ages first and put the unknown (age of the person who entered the room) as x.

14, 16, 16, 13, x

all of these numbers added to together would be:

59 + x

Because the question is talking about the mean age, to find mean:

add all the values in the data (ages) and divide by the total number of values on the list. (in your case would be 6 people (following your question) but you gave only 5 people including the x).

then (59 + x) / 6 = 21

21 x 6 = 126

126 = 59 + x

x = 67

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Consider a uniform density curve defined from x = 0 to x = 8. What percent of observations fall between 1 and 5?

allsm [11]

Answer: 0.50, which is choice b

Explanation:

The interval $1 \le x \le 5$ covers 5-1 = 4 units in the horizontal direction.

This is out of 8 units that span from x = 0 to x = 8 (we could say 8-0 = 8).

So we get the final result of 4/8 = 0.50

In other words, the interval from x = 1 to x = 5 covers exactly half of the interval from x = 0 to x = 8.

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2 years ago

The expression (3z2) (2z4) is equivalent to which of the following numerical expression? A. 6z8 B. 6z6 C. 5z6 D. 5z8

balandron [24]

B.6z6
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3 years ago

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Please help! posted picture of question

o-na [289]

C. 5x - 2y = 16 is your answer

plug in each point into the equation.

(2, -3)

x = 2, y = -3

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3 years ago

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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 k is $\frac{1}{15}$ .

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

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

Step-by-step explanation:

The random variable Y 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 k.

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

Thus, the value of k 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:

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

Check condition 1:

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

Condition 1 is fulfilled.

Check condition 2:

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

Condition 2 is not satisfied.

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

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2 years ago

If there are 16oz in a pound how many oz are in 10 pounds

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