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Westkost [7]
3 years ago
15

A toal of 21,650 balls. It sold 11,795 baseballs. It sold 4,150 fewer basketballs than baseballs. The rest of the balls sold wer

e footballs. How many did the footballs did the store sell
Mathematics
1 answer:
Lunna [17]3 years ago
4 0

Answer: 5705 Footballs

Step-by-step explanation: Just subtract 11,795 by 4,150 to get the missing variable.

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A recent survey of 8,000 high school students found that the mean price of a prom dress was $195.00 with a standard deviation of
KiRa [710]

To calculate the z-statistic, we must first calculate the standard error.

Standard error is standard deviation divided by the square root of the population. In this case, it is equal to 2.68.

The z-score is defined the distance from the sample to the population mean in units of standard error.

z = (195 – 208)/2.68 = -4.86

5 0
3 years ago
Solve - 2/5x is less than or equal to 20
sashaice [31]

Answer:

x is less than or equal to 0

Step-by-step explanation:

- you need to multiply both sides of the inequality by 5/2

- then you reduce the numbers with the greatest common factor 5

- then you reduce the numbers with the greatest common factor 2

- any expression multiplied by 0 equals 0

- so you get x is less than or equal to 0

^^^ hope this helps! :)

8 0
3 years ago
A multiple-choice standard test contains total of 25 questions, each with four answers. Assume that a student just guesses on ea
weqwewe [10]

Answer:

a)  8*88*10⁻⁶ ( 0.00088 %)

b) 0.2137 (21.37%)

Step-by-step explanation:

if the test contains 25 questions and each questions is independent of the others, then the random variable X= answer "x" questions correctly , has a binomial probability distribution. Then

P(X=x)= n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of questions= 25

p= probability of getting a question right = 1/4

then

a) P(x=n) = p^n = (1/4)²⁵ = 8*88*10⁻⁶ ( 0.00088 %)

b) P(x<5)= F(5)

where F(x) is the cumulative binomial probability distribution- Then from tables

P(x<5)= F(5)= 0.2137 (21.37%)

7 0
3 years ago
Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp
aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}

Using a statistic software I've calculated the test

F_{H_0}= 1.41

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0
3 years ago
What is 84 ft. to 84 sq ft.
svetlana [45]
12 sq. in. in a foot i think 
5 0
3 years ago
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