All categories

2 years ago

Determine whether the two ratios are proportional. 6/36 and 1/6 a) Yes b) No

1 answer:

6 0

Yes, they are. Treat ratios like fractions, simplify them the same way. In 6/36, the GCF (Greatest Common Factor) of the numerator and the denominator is six. So, we divide both sides by six. This gives us 1/6, which is still the same fraction. (Making it proportional)

You might be interested in

How much percentage of a minute is a second?

KengaRu [80]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Find the value of x.<br> A.<br> (2x-17)"<br> С

lora16 [44]

The answer would be X=34

8 0

2 years ago

An office and apartment building is made up of modules. Each module is a rectangular prism. The building has 150 modules, and is

Elis [28]

Answer:

Part 1) 10 modules per floor

Part 2) Two different rectangular prism

Step-by-step explanation:

Part 1)

Let

x ----> the number of modules

y ---> the number of stories

we have

$x=150\ modules\\y=15\ stories$

Divide the number of modules by the number of stories

$\frac{x}{y}=\frac{150}{15}=10\ \frac{modules}{floor}$

Part 2) How many different rectangular prisms could be made from that number?

The number is 10

Look at the factors of 10

They are 1,2 ,4,5 and 10

Pick any two of these factors and determine if there is a third value from this list so that the product of the three factors is 10

There are 2 combinations that will work :

1 0×1×1

2 ×5×1

4 0

3 years ago

how much does an x box emulator cost and what does it do??? Will give brainliest to first right answer!!!

ladessa [460]

Answer:

113.44

Step-by-step explanation:

It can emulate the BIOS and many games at very slow speeds but is sometimes faster than Cxbx with acceptable graphics

6 0

3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

2 years ago