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3 years ago

The constant of proportionality can be found by finding the

Mathematics

2 answers:

Amanda [17]3 years ago

8 0

The constant of proportionality can be found by finding the unit rate.

KatRina [158]3 years ago

6 0

Answer:

The constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other. Once you know the constant of proportionality you can find an equation representing the directly proportional relationship between x and y, namely y=kx, with your specific k.

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Please help will give brainliest

butalik [34]

Answer:

360 m is the answer........

6 0

3 years ago

Read 2 more answers

6. Evaluate the expression for the value of the variable. 1/3x + 5 x = -6

taurus [48]

Answer:

$\frac{1}{3} x + 5x = - 6 \\ \frac{16}{3} x = - 6 \\ x = \frac{ - 6}{ \frac{16}{3} } \\ x = \frac{ - 9}{8}$

5 0

3 years ago

Find four consecutive even integers such that seven times the first exceeds their sum by 18.

Ganezh [65]

Answer:

10, 12, 14, 16

Step-by-step explanation:

Given: Four consecutive even integers such that seven times the first exceeds their sum by 18.

Lets assume the first number be "x".

As it is even integers

∴ Second consecutive number will be $(x+2)$

Third consecutive number will be $(x+4)$

Fourth consecutive number will be $(x+6)$

Now, as given seven times the first exceeds their sum by 18.

∴ $[x+(x+2)+(x+4)+(x+6)]+18 = 7x$

solving the equation to find the number.

⇒ $[x+(x+2)+(x+4)+(x+6)]+18 = 7x$

Opening parenthesis

$x+x+2+x+4+x+6+18 = 7x$

⇒ $4x+30= 7x$

subtracting both side by 4x

⇒ $30= 3x$

Dividing both side by 3

∴ x= 10.

Hence, subtituting the value x to find four consecutive even integers.

First number is 10

Second consecutive number will be $(10+2)$ = 12

Third consecutive number will be $(10+4)$ = 14

Fourth consecutive number will be $(10+6)$ = 16

∴ Four consecutive even integers are 10,12,14,16.

8 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

3 years ago

The measure of a square wall is 9 feet 7 inches on each side. Estimate the square feet a painter would have to cover to the near

Nookie1986 [14]

First, convert those inches to feet.
7in =ish 0.5833333
(you find this by dividing 7/12 on a calculator)
Now you just square 9.583333333
9.583333333^2
9.583333333 * 9.583333333
91.84027777ish
Rounded to the nearest foot, means the answer is 92ft.

5 0

3 years ago