(I’ll give brainilest) help asap

Given: 315/1575. Write the prime fractorization of the numerator and the denominator, then write the fraction in lowest terms.

Ne4ueva [31]

The prime factorization of the numerator, 315, is 5 x 3 x 3 x 7. That of the denominator, 1575 is 5 x 5 x 3 x 3 x 7. To convert the fraction to its lowest form, cancel all the factors that appear to both the numerator and denominator. We will be left with 1/5.

3 0

3 years ago

What is The prime factorization of <br> is 24 × 5.

Anna [14]

Answer:

The prime factorization of the number 24 is 2 × 2 × 2 × 3.

4 0

2 years ago

Read 2 more answers

Brainliest goes to whoever answers correctly try to show work ONLY if you can also if you want more points answer my other quest

dolphi86 [110]

Answer:

Hi, there the answer is C. y=33/2x+425

I tried uploading a picture on how I got, but it keep saying I'm using offensive word, which I didn't but trust me the answer is right

Step-by-step explanation:

5 0

2 years ago

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:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = 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 <em>A</em>, <em>B</em> and <em>C</em> 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

A car travels at a constant speed.

ad-work [718]

Answer:

see below

Step-by-step explanation:

<h3>Given</h3>

Distance is 142.2 m, correct to 1 decimal place
Time is 7 seconds, correct to nearest second

Upper bound for the speed

<h3>Solution </h3>

<em>Upper bound for the speed = upper bound for distance/lower bound for time</em>

Upper bound for distance = 142.25 m (added 0.1/5 = 0.05)
Lower bound for time = 6.5 seconds (subtracted 1/2 = 0.5)

<u>Then, the speed is:</u>

142.25/6.5 = 21.88 m/s
21.88 = 21.9 m/s correct to 1 decimal place
21.88 = 22 m/s correct to nearest m/s

4 0

3 years ago