*starts crying from happiness*❤️❤️❤️❤️❤️❤️❤️❤️❤️

Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of

zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

$H_{0}: p = 0.29$

We want to test the hypothesis that this is true, so the alternate hypothesis is:

$H_{a}: p \neq 0.29$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that $\mu = 0.29, \sigma = \sqrt{0.29*0.71}$

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that $n = 100, X = 0.21$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}$

$z = -1.76$

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

5 0

2 years ago

Mikas diner sold 70 milkshakes last week. 9% of the milkshakes had whipped cream on top. How many milkshakes with whipped cream

VARVARA [1.3K]

Step-by-step explanation:

I hope u got ur answer ;)

if u have any more doubt ask me

8 0

2 years ago

Read 2 more answers

Which expressions are polynomials?

igor_vitrenko [27]

Option A: z + 1

Option B: 6 + w

Option D: $2 x^{4}-y$

Solution:

Let us first define the polynomial.

A polynomial can have constants, variables, exponents and fractional coefficients.

A polynomial cannot have negative exponents, fractional exponents and never divided by a variable.

To find which expressions are polynomial:

Option A: z + 1

By the definition, z + 1 is a polynomial.

It is polynomial.

Option B: 6 + w

By the definition, 6 + w is a polynomial.

It is polynomial.

Option C: $y^{2}-\sqrt[3]{y}+4$

$y^{2}-\sqrt[3]{y}+4=y^{2}-{y}^{1/3}+4$

Here, y have fractional exponent.

So, it is not a polynomial.

Option D: $2 x^{4}-y$

By the definition, $2 x^{4}-y$ is a polynomial.

It is polynomial.

Hence z + 1, 6 +w and $2 x^{4}-y$ are polynomials.

6 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .