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3 years ago

Daniel and Sofia both solve this system of equations:

Mathematics

1 answer:

yarga [219]3 years ago

6 0

You, too, can read the graph and determine that the second selection is appropriate. Or, you can graph it yourself and determine the (one) correct solution.

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Find the seventh term of the

jekas [21]

Answer:

$T_7 = \frac{64}{729}$

Step-by-step explanation:

Given

$a =1$

$r = \frac{2}{3}$

Required

Determine the 7th term

The nth term of a gp is:

$T_n = a * r^{n-1$

So, we have:

$T_7 = 1 * \frac{2}{3}^{7-1$

$T_7 = 1 * \frac{2}{3}^{6$

$T_7 = 1 * \frac{2^6}{3^6}$

$T_7 = 1 * \frac{64}{729}$

$T_7 = \frac{64}{729}$

6 0

3 years ago

use the given conditions to write a system of equations. solve the system and find the numbers. The sum of two numbers is 10. If

ratelena [41]

One of the number is 4 and the other is 6

3 0

3 years ago

A square has a perimeter of 12x+52 units. Which expression represents the side length of the square in units?

Tpy6a [65]

Answer:

C. 3x+13

Step-by-step explanation:

Because

12x+52=

4(3x+13)

The 4 is the sides of a square

8 0

3 years ago

Suppose the time to process a loan application follows a uniform distribution over the range of 8 to 13 days. What is the probab

Arlecino [84]

Answer: 0.2

Step-by-step explanation:

We know that , the probability density function for uniform distribution is given buy :-

$f(x)=\dfrac{1}{b-a}$ , where x is uniformly distributed in interval [a,b].

Given : The time to process a loan application follows a uniform distribution over the range of 8 to 13 days.

Let x denotes the time to process a loan application.

So the probability distribution function of x for interval[8,13] will be :-

$f(x)=\dfrac{1}{13-8}=\dfrac{1}{5}$

Now , the probability that a randomly selected loan application takes longer than 12 days to process will be :-

$\int^{13}_{12}\ f(x)\ dx\\\\=\int^{13}_{12}\dfrac{1}{5}\ dx\\\\=\dfrac{1}{5}[x]^{13}_{12}\\\\=\dfrac{1}{5}(13-12)=\dfrac{1}{5}=0.2$

Hence, the probability that a randomly selected loan application takes longer than 12 days to process = 0.2

8 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST

Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

$b > 3\frac{2}{15}\\\\$

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality $2 \frac{3}{5} < b - \frac{8}{15}\\\\$ is the same as $\frac{13}{5} < b - \frac{8}{15}\\\\$

---------------------

Let's multiply both sides by 15 to clear out the fractions

$\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\$

---------------------

Now isolate the variable b

$39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\$

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

7 0

3 years ago