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3 years ago

9

I have 4 green jolly ranchers and 10 blue jolly ranchers. Which best describes the likelihood of choosing a green jolly rancher?

1 answer:

ivann1987 [24]3 years ago

7 0

Answer:

4/10 simplified to 2/5

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10 points to the person who answers this whole thing and I will mark you as brainliest.

olga2289 [7]

Answer:

Step-by-step explanation:

Perimeter is found by adding together all the lengths of the sides. For us, that is x + (2x + 5) + (6x - 17) + (3x + 2). Now we will just combine like terms. We can also drop the parenthesis because they do nothing for us and mean nothing to the problem.

x + 2x + 5 + 6x - 17 + 3x + 2 becomes

12x - 10

4 0

4 years ago

P/90 = 4/18. What does p equal

Phoenix [80]

P/90 = 4/18
18P = 90*4 [cross-multiplication]
P = 360/18
P = 20
So, answer is P equals to 20

3 0

3 years ago

Read 2 more answers

Abby has a credit card which uses the adjusted balance method to compute finance charges. Her card has an APR of 11.83%, and she

levacccp [35]

The answer to this question is A: 6.08

5 0

3 years ago

Read 2 more answers

With full explanation from the internet like before<br> 3x2-6x+5=0

Elodia [21]

Answer:

$\sf x=1+i\sqrt{\dfrac{2}{3}} \ \quad and \quad \:x=1-i\sqrt{\dfrac{2}{3}}$

Explanation:

<u>Given Expression</u>:

3x² - 6x + 5 = 0

Use the Quadratic Formula:

$\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \ \ when \ \ ax^2 + bx + c = 0$

<u>insert coefficients</u>

$\Longrightarrow \sf x = \dfrac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$\Longrightarrow \sf x = \dfrac{\left6\right\pm \sqrt{-24} }{6}$

$\Longrightarrow \sf x = \dfrac{\left6\right\pm 2\sqrt{6}i}{6}$

$\Longrightarrow \sf x =1 \pm i\dfrac{\sqrt{6} }{3}$

$\Longrightarrow \sf x=1+i\sqrt{\dfrac{2}{3}}, \quad 1-i\sqrt{\dfrac{2}{3}}$

6 0

2 years ago

Read 2 more answers

Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.

Naya [18.7K]

It's a geometric sequence.

$4,-12,36,... \\ \\
a_1=4 \\
r=\frac{a_2}{a_1}=\frac{-12}{4}=-3 \\ \\
a_n=a_1 \times r^{n-1} \\
a_n=4 \times (-3)^{n-1} \\
a_n=4 \times (-3)^{-1} \times (-3)^n \\
a_n=4 \times (-\frac{1}{3}) \times (-3)^n \\
a_n=-\frac{4}{3}(-3)^n$

It's the sum for term 4 through term 15.

$\boxed{ \sum\limits_{n=4}^{15} (\frac{4}{3}(-3)^n)}$

8 0

3 years ago