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3 years ago

10

A box contains 15 yellow, 5 orange, and 15 green tennis balls. If Izzy draws a tennis ball at random out of the box what is the

probability that she drew either a green or an orange tennis ball?

2 answers:

kumpel [21]3 years ago

8 0

Answer:

Green= 15/35 0r 3/7 Orange= 5/35 0r 1/7 Both= 20/35 or 4/7

Step-by-step explanation:

Add up all the balls = 35

Add the orange and green balls = 20

Put into fraction.

20 balls out of the 35.

20/35 or 4/7

Hope this helps.

Eva8 [605]3 years ago

4 0

Answer:

P(green or orange) = 4/7

Step-by-step explanation:

We need to know how many tennis balls there are

total = 15 yellow+ 5 orange+ 15 green

= 35 balls

Probability of green or orange tennis ball is the sum of the green and orange tennis balls over the total balls

P(green or orange) = (green + orange)/total

= (15+5)/35

=20/35

=4/7

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Ethan's full backpack weighed too much. He took out a 4/8 pound electronic reader, a 2 ⅛ pound full bottle of water, and a ⅞ pou

Luba_88 [7]

Answer:

see below

Step-by-step explanation:

We can just add up all of these fractions:

2 1/8 + 7/8 = 2 8/8 = 3

8 5/8 + 4/8 = 8 9/8 = 9 1/8

3 + 9 1/8 = $12\frac{1}{8}$

4 0

3 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago

Find a solution of the linear inequality

evablogger [386]

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3 0

4 years ago

Graph the function f(x) = x^3+2x^2-3

Ivan

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5 0

3 years ago

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

stiks02 [169]

Answer:

$0.0497 - 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.026$

$0.0497 + 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.073$

The 90% confidence interval would be given by (0.026;0.073)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated proportion for this case is:

$\hat p =\frac{322-306}{322}= 0.0497$

Represent the proportion of defectives for this case

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.0497 - 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.026$

$0.0497 + 1.96\sqrt{\frac{0.0497(1-0.0497)}{322}}=0.073$

The 90% confidence interval would be given by (0.026;0.073)

3 0

3 years ago