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4 years ago

Is 3x^2 − y = 5 a function or not a function

Mathematics

1 answer:

kakasveta [241]4 years ago

8 0

hi!

its not a function since it has a =5

this is an equation

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Della by tree seedling that was 2 1/4 feet tall during the first year it grew 1 1/6 ft after two years it was 5 ft tall how much

jasenka [17]

<h2>Greetings!</h2>

Answer:

$\frac{19}{12}$ feet

Step-by-step explanation:

First, we need to find the height after one year:

2.25 + 1 1/6 = $\frac{41}{12}$

Now to find the amount in the second year, we can subtract the above amount from 5:

5 - $\frac{41}{12}$ = $\frac{19}{12}$

So in the second year it grew $\frac{19}{12}$ feet

<h2>Hope this helps!</h2>

8 0

3 years ago

What is the area of a square, in square feet, whose perimeter is 20 feet

geniusboy [140]

Perimeter of a square = 20 feet = 4a
=>4a = 20
=>a = 20/4
=>a = 5 feet
Area of a square = a²
=>a² = 5²
=>a = 25 square feet

3 0

4 years ago

HELPPP PLEASE!! Which two fixed points define the shape ellipse

Marysya12 [62]

Answer: The answer is b i took the test plsss tell me if it helps;)

8 0

3 years ago

A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed f

Olin [163]

Answer:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the frequencies

H1: There is a difference in the frequencies

The level of significance assumed for this case is $\alpha=0.01$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are:

Value 1 2 3 4 5 6

Frequency 26 32 44 37 27 34

And the expected values are for this case the same $E_i = \frac{200}{6}= 33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

7 0

4 years ago

Mr. And Mrs. Jones wanted to plan a fun day with their two children. An adult ticket to the local amusement park is $20. A child

Levart [38]

Answer:

The total cost will be $60 dollars.

Step-by-step explanation:

If the adults tickets are $20 and the children's are 50% less

$20 × 0.5 = $10

so with two adults and two kids

20 + 20 + 10 + 10 = $60

3 0

3 years ago