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3 years ago

A bag contains 5 red marbles, 6 blue marbles and 9 yellow marbles.

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Answer:

Step-by-step explanation:

The total number of marbles is 20

Question A

P(Red) = 4/20 = 1/5

P(yellow) = 9/20

P(Red then Yellow) = 1/5 * 9/20 = 9 / 100

Question B

P(B) = 6/20

P(B again) = 5/19 because you have 1 less total plus 1 less blue.

P(B and B) = 6/20 * 5/19

P(B and B) = 30/ 380 = 3/38

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What property is shown here? (1 x 4) x 3 = 1 x (4 x 3)

irinina [24]

Answer:Associative Property of Multiplication

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3 years ago

Angela is putting a fence around her garden, which is bordered by her house on one side (as shown in the picture).

joja [24]

Answer:

B:5x

Step-by-step explanation:

It would be B because if you add all the sides together you would get 5x and 5x is B

5 0

3 years ago

If EF = 2x-7, FG = 4x-20, and EG = 21, find the values of x, EF, and FG.

Serggg [28]

Given -:

EF = 2X-7

FG = 4X-20

EG = 21

FIND OUT

EF =?

FG = ?

To proof

as given in the question

we have the value of EF = 2X-7 , FG = 4X-20 and EG = 21

Thus we have

EG = EF + FG

21 = 2X-7 + 4X- 20

21 = 6x - 27

48 = 6x

x = 8

Hence proved

Now put this value

$EF = 2\times 8 - 7\\EF = 9$

Now

$FG = 4\times 8 - 20\\FG = 12$

Hence proved

8 0

3 years ago

Help PLSSSSSSS If you can now

marin [14]

Answer:

its the 3rd option not the second

Step-by-step explanation:

4 0

2 years ago

Toilet Training You are a great friend and are taking your friend's child to the playground to play with two of his friends: Cha

Brilliant_brown [7]

Answer:

Step-by-step explanation:

Hello!

According to a study regarding the average age of female and male kids to complete toilet training is:

Females:

Average age 32.5 months

The standard deviation of 6.7 months

n= 126

Males:

Average age 35.0 months

The standard deviation of 10.1 months

n= 141

The parameter of study is the difference between the age of females are successfully toilet trained and the average age that males are successfully toilet trained. μf - μm (f= female and m= male)

Assuming that both variables have a normal distribution to choose whether you'll use an unpooled or pooled-t to calculate the confidence interval you have to conduct an F-test for variance homogeneity.

If the variances are equal, then you can usee the pooled-t, but if the variances are different, you have to uses Wlche's approach:

H₀: δ²f = δ²m

H₁: δ²f ≠ δ²m

Since both items b. and c. ask for a 95% CI I'll use the complementary significance level for this test:

α: 0.05

$F= \frac{S^2_f}{S^2_m} * \frac{xSigma^2_f}{Sigma^2_m} ~~~F_{(n_f-1); (n_m-1)}$

$F= \frac{(6.7^2)}{(10.1)^2} *1= 0.44$

Critical values:

$F_{125;140;0.025}= 0.71\\F_{125;140;0.975}= 1.41$

The calculated F value is less than the lower critical value, 0.77, so the decision is to reject the null hypothesis. In other words, there is no significant evidence to conclude the population variances of the age kids are toilet trained to be equal. You should use Welch's approach to construct the Confidence Intervals.

$Df_w= \frac{(\frac{S^2_f}{n_f} + \frac{S^2_m}{n_m} )^2}{\frac{(\frac{S^2_f}{n_f} )^2}{n_f-1} +\frac{(\frac{S^2_m}{n_m} )^2}{n_m-1} }$

$Df_w= \frac{(\frac{6.7^2}{126} + \frac{10.1^2_m}{141} )^2}{\frac{(\frac{126^2}{126} )^2}{126-1} +\frac{(\frac{10.1^2}{141} )^2}{141-1} } = 254.32$

The given interval is:

[0.3627; 4.6073]

Using Welch's approach, the formula for the CI is:

(X[bar]f- X[bar]m) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

(X[bar]m- X[bar]f) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

As you can see either way you calculate the interval, it is centered in the difference between the two sample means, so you can clear the value of that difference by:

(Upper bond - Lower bond)/2= (4.6073-0.3627)/2= 2.1223

The average age for females is 32.5 months and for males, it is 35 months.

Since the difference between the sample means is positive, we can say that the boys were considered "group 1" and the girls were considered "group 2"

You have a95% confidence that the parameter of interest is included in the given confidence interval.

None of the statements is correct, the interval gives you information about the difference between the average age the kids are toilet trained, that is between the expected ages for the entire population of male and female babies.

This represents a guideline but is not necessarily true to all individuals of the population since some male babies can be toilet trained before that is expected as some female babies can be toiled trained after the average value.

I hope it helps!

6 0

3 years ago