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3 years ago

A 6 foot tall man makes a shadow that is 3 feet long. How tall is a building that

Mathematics

1 answer:

Allisa [31]3 years ago

4 0

Answer:

A) 1 : 12/7

B) see explanation below

C) 25.5 feet

Step-by-step explanation:

6 : 3.5 --> 1 : 12/7

x : 14.875 --> 1 : 12/7

x= (14.875) (12/7)

x= 25.5

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Benjamin writes an expression for the sum of 1 cubed, 2 cubed and 3 cubed. What is the value at the expression

asambeis [7]

Answer:

Step-by-step explanation:

8 0

3 years ago

−1/2÷7 1/2 dividing fraction

harina [27]

-0.0375

-1/2
———- ( 7 1/2 ) =-3/80
100

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3 years ago

Solve the simultaneous equations y=3x+2 and y=x+6

kogti [31]

Answer:

x=2

Step-by-step explanation:

y=3x+2

y=x+6

Since you're stating that, they both equal because equation is equated to equation 2 because of the Y.

Therefore, substitute into both equations as shown:

3x+2=x+6

3x-x=6-2 (bring 2 to the other side by subtracting 2 as well as x)

2x=4

2x/2=4/2 (dividing 2 on both sides to get rid off 2 in 2x)

x=2.

5 0

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.

7 0

3 years ago

What is the common denominator for 32 and 3

Leya [2.2K]

Just multiply 32 by 3

4 0

3 years ago