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3 years ago

9

A box contains 6 red marbles and 5 blue marbles. another box has 5 red and 8 blue marbles. One box is selected at random and fro

m that box one marble is drawn. What is the probability that the marble drawn is blue

1 answer:

irga5000 [103]3 years ago

3 0

Most likely uncertain because it is a small amount and there are bigger amounts.

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The average score on a standardized test is 750 points with a standard deviation of 50 points. If 2,000 students take the test a

tatuchka [14]

I the average score is 750 and standard deviation ( sigma ) is 50 points, then score 700 - 800 is between (750 - 1 sigma) and (750 + 1 sigma).
That is 34.1 % + 34.1 % = 68.2 % of the whole population.
2,000 · 0.682 = 1.364
We expect 1,364 students to score between 700 and 800 points.

6 0

3 years ago

The mean computed from ungrouped data is a more accurate measure than the mean computed from grouped data?

Rufina [12.5K]

It doesn’t make any difference whether or not the data is grouped, the mean is the same. So b. false.

4 0

3 years ago

Determine the common ratio and find the next three terms of the geometric sequence 9,3sqrt3,3

Maru [420]

Answer:

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$

Step-by-step explanation:

The geometric progression is:

$9, 3 \sqrt{3}, 3...$

The first term, a, is 9

To find the common ratio, r, all we have to do is divide a term by its preceding term.

Let us divide the second term by the first:

$r = \frac{3\sqrt{3}}{9}\\ \\r = \frac{\sqrt{3}}{3}$

That is the common ratio.

Geometric progression is given generally as:

$a_n = ar^{(n - 1)}$

where a = first term

r = common ratio

$a_n$ = nth term

We need to find the 4th, 5th and 6th terms.

Fourth term: $a_4 = 9 * (\frac{\sqrt{3}}{3})^{(4 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{3}$ = $\sqrt{3}$

Fifth term: $a_5 = 9 * (\frac{\sqrt{3}}{3})^{(5 - 1)}$ = $9 * (\frac{\sqrt{3}}{3})^{4}$ = 1

Sixth term: $a_6 = 9 * (\frac{\sqrt{3}}{3})^{(6 - 1)} = 9 * (\frac{\sqrt{3}}{3})^{5} =\frac{\sqrt{3}}{3}$

5 0

3 years ago

Does anyone know the answer to this

Tcecarenko [31]

Answer:

9/-12

Step-by-step explanation:

add and pay attention to tye problem it is pretty self explanatory rise over run help here tho

7 0

2 years ago

Any number that is divisible by 3 also divisible by 6. Find a counterexample to show that the conjecture is false.

Alex_Xolod [135]

Answer:

Any number that is divisible by 3 is also divisible by 6. Find a counterexample to show that the conjecture is false. 24 18 12 21.

Step-by-step explanation:

brainlies plssssssss!

5 0

3 years ago