All categories

2 years ago

Determine the solutions to the following equation.

Mathematics

2 answers:

guajiro [1.7K]2 years ago

5 0

The answer would be A) x = -5 and 13.

Hope this helps :)

densk [106]2 years ago

5 0

Answer:

A- -5 and 13

You might be interested in

The average score of this quiz is 24 points with a standard deviation of 1.5 points. What is the probability that someone scored

navik [9.2K]

Answer:

8.08 %

Step-by-step explanation:

Let us denote : $X=$ score X follows a normal distribution with mean $=24$ and standard deviation $=1.5$

Step 2

The probability that someone scored 21.9 or less is :

$\mathrm{P}(\mathrm{X} \leq 21.9)$

$=\mathrm{P}\left(\mathrm{Z} \leq \frac{21.9-24}{1.5}\right)$

$=\mathrm{P}(\mathrm{Z} \leq-1.4)$

$=0.0808$

$=8.08%$

Answer : $8.08 %$

7 0

2 years ago

Solving quadratics equations by taking square root

Novay_Z [31]

#6
x^2 = 1
x = - 1 and x = + 1

#7
a^2 + 1 = 19
a^2 =18
a^2 =√ 9 √ 2
a = + 3√ 2 and a = - 3√ 2

4 0

3 years ago

Samantha works at an animal shelter. She decides to track the total number of kittens that are brought in. The

Softa [21]

Answer:

C) The domain represents the weeks that have passed since Samantha started counting the kittens. The domain is all whole numbers.

Step-by-step explanation:

The problem statement tells you the independent variable w represents weeks that have passed. "Domain" refers to values the independent variable may have, so choices A or B make no sense here.

Time is measured continuously, and fractions of a week are possible. So, the domain could be non-negative real numbers. However, the answer choice D is "all real numbers", which includes negative numbers for which the function makes no sense.

The domain "all whole numbers" includes non-negative integers. It is reasonable to restrict the domain to non-negative integer numbers of weeks, so answer choice C is the best option.

8 0

3 years ago

In tossing four fair dice, what is the probability of tossing, at most, one 3

11Alexandr11 [23.1K]

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )