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3 years ago

11

Josh estimates the height of his desk. What is a reasonable estimate?

1 answer:

kaheart [24]3 years ago

5 0

3ft atleast or 2 and a half feet

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Please help I cannot figure this one out!

ra1l [238]

Answer:

77.47%

Step-by-step explanation:

We have to use Bayes Theorem in solving this problem.

This theorem's formula is:

P(A | B) = $\frac{P(A and B)}{P(B)}$

Where the " | " means <u><em>"given that"</em></u>.

For our problem, using bayes theorem, we can write:

P(Student has lice | positive ) = $\frac{P(StudentHasLice and Positive)}{P(Positive)}$

P(Positive) = 0.1692 + 0.0492 = 0.2184
P (Student has Lice & Positive ) = 0.1692

So P(Student has lice | Positive ) = $\frac{0.1692}{0.2184}=0.7747$

4 0

3 years ago

Write an inequality.<br> The sum of a number and three is no more than eight or is more than twelve.

viktelen [127]

Answer:

8 ≥ x+3 or x ≥ 12

Step-by-step explanation:

The sum of a number (additon) and 3 (x+3) is no more than (so less than or equal to eight) OR is more than 12 ( x is more than 12, x ≥ 12 )

8 0

3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago

Solve the fraction <br>1.2 1/2-1 1/4<br>2.1/2+1/4<br>3.1/3+2/5<br>4.1/2-1/3<br>5.1/4-1/3

DanielleElmas [232]

Answer:

Step-by-step explanation:

1. 1.25

2. 0.75

3.0.73333333333

4. 0.16666666666

5. 0.083333333333333

3 0

3 years ago

In largo city, about 1 in every 9 homes contain 4 televisions

schepotkina [342]

Answer:

1/9 9:1 9 to 1

1/4 1:4 1 to 4

Step-by-step explanation:

5 0

3 years ago

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