1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aivan3 [116]
3 years ago
5

Ellen is opening. Cookie shop eat a local middle school . She randomly surgery's students to determine The types of cookies they

would buy at the cookie shop the results of the surveys are below
Mathematics
1 answer:
Ket [755]3 years ago
6 0

Answer:

The only statement that is NOT true is option A.

- If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

Step-by-step explanation:

Complete Question

Ellen is opening. Cookie shop eat a local middle school . She randomly surgery’s students to determine The types of cookies they would buy at the cookie shop the results of the surveys are below based on the survey. Which statement is not true?

Cookie Survey

Type of Cookie | Number of Students

Chocolate Chip | 25

Oatmeal Raisins | 15

Peanut Butter | 22

Sugar | 18

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

B) 22½% of the students surveyed would buy a sugar cookie.

C) If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

Solution

First of, we obtain the proportion or percentages of students surveyed that buy each type of cookie.

Each of that is the number of students that buy the type of cookie in the sample divided by the Total number of students sampled.

Total number of students sampled = 25 + 15 + 22 + 18 = 80

Type of Cookie | N | proportion | %

Chocolate Chip | 25 | 0.3125 | 31.25%

Oatmeal Raisins | 15 | 0.1875 | 18.75%

Peanut Butter | 22 | 0.2750 | 27.50%

Sugar | 18 | 0.2250 | 22.50%

Now, taking the statements one by one

A) If 160 students buy a cookie, approximately 18 students would buy a sugar cookie.

Proportion of students that buy sugar cookie in the sample = 0.225

Number of students that will buy sugar cookie amongst 160 students = np = 160 × 0.225 = 36 students.

This statement is false, that is, NOT true.

B) 22½% of the students surveyed would buy a sugar cookie.

From the table, 22.5% of the students surveyed bought a sugar cookie. Hence, no matter the number of students, the percentage of students that will buy a sugar cookie will approximately be 22.5%.

This statement is True.

C) If 240 students were to purchase a cookie from the store, approximately 66 students will purchase a peanut butter cookie.

Proportion of students that buy peanut butter cookie in the sample = 0.275

Number of students that will buy peanut butter cookie amongst 240 students = np = 240 × 0.275 = 66 students.

This statement is true.

D) Half of the students prefer chocolate chips or oatmeal raisin cookies.

Proportion of students that buy chocolate chips or oatmeal raisin cookies in the sample = 0.3125 + 0.1875 = 0.50

0.50 is half of the distribution.

Hence, this statement too, is True.

Hope this Helps!!!

You might be interested in
Jerry multiplied 3.04 x 7.091 and got the product 2.155664. Is his answer reasonable? Why or why not?
Readme [11.4K]

First we multiply the numbers ignoring the decimal and then place the decimal in the position

3.04 *7.091 = 2155664

Now the first number has two digits after the decimal and the second number has three digits after the decimal.

Therefore in the final result we have to move the decimal (2+3) =5 place from the left,

So the answer would be 21.55664

So, the correct answer is 21.55664

Jerry calculated the answer correctly , but made an error in the placement of the decimal point.  The answer is not reasonable because 2.155664 and 21.55664 mean two completely different answer far apart from each other


6 0
2 years ago
(17x + 21)<br> 123<br> need the answer asap
Greeley [361]
What are we trying to solve?
7 0
3 years ago
Read 2 more answers
Please solve this simultaneous equation
Scrat [10]

Answer:

b6 2hich method

Step-by-step explanation:

I solv3 this

3 0
3 years ago
Read 2 more answers
What is wrong with the following proof that for every integer n, there is an integer k such that n &lt; k &lt; n + 2? Suppose n
expeople1 [14]

Answer:

c)The proof writer mentally assumed the conclusion. He wrote "suppose n is an arbitrary integer", but was really thinking "suppose n is an arbitrary integer, and suppose that for this n, there exists an integer k that satisfies n < k < n+2." Under those assumptions, it follows indeed that k must be n + 1, which justifies the word "therefore": but of course assuming the conclusion destroyed the validity of the proof.

Step-by-step explanation:

when we claim something as a hypothesis we can only conclude with therefore at the end of the proof. so assuming the conclusion nulify the proof from the beginning

4 0
3 years ago
Trouble finding arclength calc 2
kiruha [24]

Answer:

S\approx1.1953

Step-by-step explanation:

So we have the function:

y=3-x^2

And we want to find the arc-length from:

0\leq x\leq \sqrt3/2

By differentiating and substituting into the arc-length formula, we will acquire:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx

To evaluate, we can use trigonometric substitution. First, notice that:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx

Let's let y=2x. So:

y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx

We also need to rewrite our bounds. So:

y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0

So, substitute. Our integral is now:

\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Let's multiply both sides by 2. So, our length S is:

\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

y=a\tan(\theta)

Substitute 1 for a. So:

y=\tan(\theta)

Differentiate:

y=\sec^2(\theta)\, d\theta

Of course, we also need to change our bounds. So:

\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0

Substitute:

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta

The expression within the square root is equivalent to (Pythagorean Identity):

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta

Simplify:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta

And:

dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)

Integration by parts:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)

Distribute:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)

Now, let's make the single integral into two integrals. So:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Distribute the negative:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Divide the second and third equation by 2. So: \displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))

Therefore, our arc length will be equivalent to:

\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Divide both sides by 2:

\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Evaluate:

S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))

Evaluate:

S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))

Simplify:

S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}

Use a calculator:

S\approx1.1953

And we're done!

7 0
3 years ago
Other questions:
  • Tim walked 2/5 of a mile on Monday, 6/7 of a mile on Tuesday, and 2/3 of a mile on Wednesday. How far did he walk in all?
    11·2 answers
  • Can someone answer this what is 3/4 times 3/6 in simpilest form
    7·2 answers
  • Suppose a movie theater you work at sells popcorn in three different sizes. A small costs $2, a medium costs $5, a large costs $
    5·1 answer
  • Choose an equation in slope-intercept form for the line that passes through (0,7) and is parallel to the y=4x-19
    9·2 answers
  • Write an expression, using an exponent, that is equivalent to 9\times 9\times 9\times 9\times 9\times 9\times 99×9×9×9×9×9×9
    6·2 answers
  • Solve the system by the substitution method. Show your work.<br> 2y - x = 5<br> x^2 + y^2 - 25 = 0
    9·1 answer
  • JoAnn works in a publicity office at the state university. She is paid $13.50 an hour for the first 35 hours she works each week
    7·2 answers
  • Can some one help me please god bless you <br><br><br> 25 points
    12·1 answer
  • why do u get to determain how many daily limits i can have u encourge alot of a ton people to buy you're vip for over priced pri
    6·1 answer
  • Help me please they are going to kick me out if not
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!