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3 years ago

Find the Value of X. If your answer is not an integer, express it in simplest radical form.

Mathematics

1 answer:

Gennadij [26K]3 years ago

6 0

For each question we must apply the Pythagorean theorem and clear the value of x.
question 5
x = root (14 ^ 2 + 16 ^ 2)
x = root (452)
x = 2raiz (113)
question 6
x = root (8 ^ 2 + 15 ^ 2)
x = 17
question 7
x = root (12 ^ 2 + 12 ^ 2)
x = root (2 * (12 ^ 2))
x = 12raiz (2)
question 8
x = root (18 ^ 2 - 9 ^ 2)
x = root (243)
x = 9raiz (3)

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a rental car agency charges a flat fee of $110 plus $46 per day to rent a certain car. another agency charges a fee of $70 plus

insens350 [35]

First make and equation. Since you want to find the number of the day where both charge the same total amount, you would make the equation:
let x= the number of days
110+46(x)=70+54(x)
Solve:
110-70+46(x)=70-70+54(x)
40+46(x)=54(x)
40+46(x)-46(x)=54(x)-46(x)
40=8(x)
40/8=8x/8
5=x
Therefore your answer would be 5 days

8 0

3 years ago

Read 2 more answers

813 divided by 4 and find remainder

Artemon [7]

203 r 25 would be the answer

8 0

2 years ago

Read 2 more answers

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0

3 years ago

The decimal 0.26 as a fraction

Nitella [24]

Simplified: 13/50
Not Simplified: 26/100

4 0

3 years ago

Is the data set approximately periodic?

wolverine [178]

Answer:

Option: C is the correct answer.

C. Periodic with period 4 and amplitude of about 15.

Step-by-step explanation:

We are given the table of values as:

Hour 1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9 -- 10 -- 11

Number 24-41-10-32-26-38-9-31-25-39-12

of boats

As we could see that after every 4 value the values are approximately repeating.

Hence,the data set is approximately periodic.

Also, the amplitude is calculated as:

Average of the maximum and minimum value of each set

We divide the set into 3 sets as follows:

Hour 1 2 3 4

Number of boat 24 41 10 32

Here we have amplitude as:

(41-10)/2=15.5

Hour 5 6 7 8

Number of boat 26 38 9 31

Amplitude is:

(38-9)/2=14.5

Hour 9 10 11 12

Number of boat 25 39 12

Amplitude is:

(39-12)/2=13.5

Hence, we see that the amplitude is about 15.

4 0

3 years ago