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3 years ago

What is the longest line segment that can be drawn in a right rectangular prism that is 13 cm long,

Mathematics

2 answers:

Ganezh [65]3 years ago

7 0

Step-by-step explanation:

The line segment (or chord) AB (in the figure), which passes through the foci (F1, F2) and terminates on the ellipse, is called the major axis. This axis is the longest segment that can be obtained by joining two points on the ellipse. The two points at which the major axis intersects the curve are called the vertices.

givi [52]3 years ago

5 0

Answer:

15.81 cm

Step-by-step explanation:

The diagonal line that links the 13 cm and 9 cm sides will be longest.

Use the Pyth Theorem to find it:

13^2 + 9^2 = c^2

169 + 81 = c^2

250 = c^2

15.81 = c

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What is the value of s in the equation 3 r equals 10 plus 5 s, when r equals 10?

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Answer:

S=4

Step-by-step explanation:

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3 years ago

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A gas tank has a leak. Each hour it loses 3 gallons of gas. Write the equation that represents how much gas, g , the tank loses

zhenek [66]

Let t is the amount of time that the gas is leaking (hour), g is the total of the gas leak.

g = 3t.

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3 years ago

The members of a club decide to sell hats to raise money. They originally

finlep [7]

Answer:

A. 41 (12 - 2)

Step-by-step explanation:

If the club decided to charge $12 for a hat, but reduced the price by $2, the reduced price of the hat is (12 - 2). The amount of money earned is based upon the price of the hat multiplied by the number of hats sold. Therefore, the answer is 41 (12 - 2).

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2 years ago

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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3 years ago

Solve for p P-17=2p+3

Y_Kistochka [10]

Answer:

-20

Step-by-step explanation:

P-17=2P+3

P-2P=17+3

-P=20

Divide both sides by negative 1

P=20

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4 years ago