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3 years ago

For which equation is x = 1 a solution?

Mathematics

1 answer:

andreev551 [17]3 years ago

3 0

Answer:

C x+3=1 is the answer to your question

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What is the quadratic regression equation for the data set?

Svet_ta [14]

Answer:

y=−0.004x2+0.415x−0.546

6 0

3 years ago

How are rational exponents and radicals related? Summarize how to rewrite a rational power as a radical term using an example of

Nataly [62]

<span>Rational exponent are exponents written in the form of a fraction.
A rational power can be rewritten as a radical in the following way: the bottom of the rational exponent is the root, while the top of the rational exponent is the new exponent on the radical.
Example: X^(1/2)= </span>√X

3 0

3 years ago

Write the first four terms of the sequence:<br> a = 2<br> an = a(n-1) + 4

eduard

We have sequence equation $a_n=a(n-1)+4$ .

In this case n is a natural number (1, 2, 3, ...).

So start inserting and computing value of $a_n$ given that you know the value of n and a.

$a_1=2(1-1)+4=4$ (first term)

$a_2=2(2-1)+4=6 \\a_3=2(3-1)+4=8 \\a_4=2(4-1)+4=10$

Hope this helps.

3 0

2 years ago

A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla

nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = $P(A)= \frac{13}{100}=0.13$

The probability that athlete is a basketball player = $P(B)= \frac{52}{100}=0.52$

The probability that athlete is both basket ball player and football player = $P(A\cap B) = \frac{9}{100}=0.09$

We have to find the probability that an athlete chosen is either a football player or a basketball player $P(A\cup B)$ .

Now we know that;

$P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%$

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B2%7D%7B2%7D" id="TexFormula1" title="\sqrt{\frac{2}{2}" alt="\sqrt{\frac{2

klemol [59]

2/2 is 1
Square root of 1 is 1

4 0

3 years ago