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1 year ago

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Of the 100 students enrolled at the dance studio, 50 take ballet and 24 take tap dancing. Eight of the students who take ballet

also take tap dancing. If a student at the studio is chosen at random, find the probability that they take ballet, if it is known that they do not take tap dancing

2 answers:

Soloha48 [4]1 year ago

8 0

Answer:

42 over 100

Step-by-step explanation:

MrRa [10]1 year ago

6 0

Answer:21/38

Step-by-step explanation:

You might be interested in

A cash register contains q quarters. Write an algebraic expression that represents the total amount of money (in dollars).

kati45 [8]

Answer:

0.25q

Step-by-step explanation:

A cash register is a machine used in small business that sums, display and records the total amount made by each sale. A cash register also has a drawer which is used to store money.

A quarter is a coin in the United States which is 25 cents.

Converting 25 cents to dollars:

100 cents = $1

25 cents = 25 cents × $1 / 100 cent = $0.25

The total amount of money registered = number of quarters × $0.25

The total amount of money registered = q × $0.25 = 0.25q

5 0

1 year ago

Which of the following is an example of a null hypothesis? A. The average quantity of detergent put into a box by this filling m

gavmur [86]

Answer:

C

Step-by-step explanation:

Null hypothesis: hypthesis to test that there is no significant difference between the specific characteristic of a population. Analysts look to reject a null hypothesis

A. the shipping company's average delivery time is different from 3 days. This is an example of alternative hypothesis. Null hypothesis is writtien as a claim

B. This again is an example of alternate hypthesis. The claim that mean is 0.03 is rejected with the results

C. This is a claim

D. This is rejection of a claim that mean is 1 pound

E. This is rejection of claim that average delivery time is 3 days.

5 0

1 year ago

Find the eccentricity, b. identify the conic, c. give an equation of the directrix, and d. sketch the conic.

densk [106]

Answer:

a) 10/3

b) hyperbola

c) x = ± 6/5

Step-by-step explanation:

a) A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:

$r=\frac{ep}{1\pm e*cos\theta}$

Given the conic equation: $r=\frac{12}{3-10cos\theta}$

We have to make it to be in the form $r=\frac{ep}{1\pm e*cos\theta}$ :

$r=\frac{12}{3-10cos\theta}\\\\multiply\ both\ sides\ by\ \frac{1}{3} \\\\r=\frac{12*\frac{1}{3}}{(3-10cos\theta)*\frac{1}{3}}\\\\r=\frac{12*\frac{1}{3}}{3*\frac{1}{3}-10cos\theta*\frac{1}{3}}\\\\r=\frac{4}{1-\frac{10}{3}cos\theta } \\\\r=\frac{\frac{10}{3}(\frac{6}{5} ) }{1-\frac{10}{3}cos\theta }$

Comparing with $r=\frac{ep}{1\pm e*cos\theta}$

e = 10/3 = 3.3333, p = 6/5

b) since the eccentricity = 3.33 > 1, it is a hyperbola

c) The equation of the directrix is x = ±p = ± 6/5

6 0

1 year ago

Sameera predicted that she would sell 38 blankets, but she actually sold 28 blankets. Which expression would find the percent er

antoniya [11.8K]

Solution: We are given:

Predicted Sales by Sameera $=38$

Actual Sales by Sameera $=28$

Now to find the Percent error, we have to use the below formula:

$Percent-Error= \frac{|Predicted-value - Actual-value|}{Actual-value} \times 100 \%$

$=\frac{|38-28|}{28} \times 100 \%$

$=\frac{10}{28}\times 100 \%$

$=35.71 \%$

Therefore, the percent error is $35.71 \%$

8 0

2 years ago

Read 2 more answers

A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot

Damm [24]

Answer:

a) $p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) $\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.

Step-by-step explanation:

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".

Part a

Description in words of the parameter p

$p$ represent the real population proportion of people who showed partial or complete resistance to the antibiotic

$\hat p$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

$z_{\alpha/2}$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{973}{1714}=0.568$ represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545$

$0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591$

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.

5 0

1 year ago