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4 years ago

5

A bag contains 4 green marbles, 3 blue marbles, and 8 red marbles. What is the probability that you will draw a blue marble when

randomly selecting one marble from the bag?
a. 1/4
b. 1/5
c. 2/12
d. 1/3

2 answers:

Romashka-Z-Leto [24]4 years ago

7 0

Answer:

1/5

Step-by-step explanation:

The total number of marbles in the bag is fifteen, and the total number of blue marbles is 3/15. 3/15 simplified is 1/5.

swat324 years ago

3 0

Answer:

1/5 is right one

Step-by-step explanation:

P = blue marble / total

=3/15

=1/5

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A rectangular field is 15 yards in length and 66 feet width. What is the area of the field in square feet and the area of the fi

kumpel [21]

Swap yards to feet 3 feet= 1 yard so you end up with 45 feet x 66 feet at that is 2,970 so you will end up with 2,970 feet in area and 35,640 inches.

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4 years ago

In a 30-60-90 triangle, if the the side opposite 30 degrees has a length

son4ous [18]

Might Want to draw this and Its very long but Hope it helped :)

Answer:

The sides are all of the same length - let's say a. The angles are all the same too, and since the angles must add up to 180∘, we conclude that the three angles in the equilateral triangle are equal to 180∘/3=60∘.

Now we do something sneaky. We draw a line all the way down from the top vertex of the triangle to the midpoint of the bottom line.

This new line cuts our equilateral triangle in half. What are the angles in one half?

The angle at the bottom is 90∘.

One of the angles is the same as one of the angles in the original equilateral triangle, so it is 60∘.

So the third angle must be 180∘−90∘−60∘=30∘.

Now the hypotenuse of this new triangle is a, the side length of the equilateral triangle. And the length of the shortest side is a/2, since the line we drew cut the bottom line in half.

8 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = <em>mean age of student cars.</em>

$\mu_2$ = <em>mean age of faculty cars.</em>

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, <u><em>the test statistics</em></u> = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago

Simplify: 4 - 5.2 + 6 + 7.9 - 8

eimsori [14]

3.9x+0.8y is the answer to your question.

3 0

3 years ago

Write a expression for 18.68+11.72+19.76+18.68

Ilia_Sergeevich [38]

<u>ANSWER: </u>

The simplified form of the given expression 18.68+11.72+19.76+18.68 after applying BODMAS rule is 4684.

<u> SOLUTION: </u>

The expression given to us is <em>18.68+11.72+19.76+18.68.</em> We are to simplify the equation.

To simplify the equation we have to follow the BODMAS rule.

Which sets different mathematical operations at different levels of priority.

BODMAS mean brackets, of, division, multiplication, addition and subtraction.

To solve different mathematical equations we have to follow this order of priority to get the correct answer.

Now applying BODMAS rule to the given expression we have to solve it to get an answer.

18.68+11.72+19.76+18.68

As the rule states we have to do multiplication before addition. Therefore,

1224+792+1444+1224

Now pursuing the equation for addition we get

2016+2668=4684

Hence the simplified form of the given expression after applying BODMAS rule is 4684.

3 0

3 years ago