Ask question

Ask question

All categories

3 years ago

13

Alana and Ramiro are writing an article for the school newspaper about whether their classmates prefer to buy their lunch or bri

ng their lunch from home. Alana surveyed every third student who entered the lunchroom and determined that 53% of students prefer to bring lunch from home. Ramiro surveyed every other student in line to buy lunch, and found that only 14% of students prefer to bring their lunch from home.

2 answers:

MrRissso [65]3 years ago

3 0

Answer:Ramiro’s method is

Step-by-step explanation:

Did the assignment

Eva8 [605]3 years ago

3 0

Answer:

ramiros methos is best

Step-by-step explanation:

You might be interested in

All you need is in the photo <br><br><br>please explain step by step

V125BC [204]

Answer:

Hi there

The formula is

A=p (1+r)^t

A future value

P present value

R interest rate

T time

A) A=2,000×(1+0.04)^(3)=2,249.728

B) A=2,000×(1+0.04)^(18)=4,051.63

C) 2500=2000 (1+0.04)^t

Solve for t

T=log(2,500÷2,000)÷log(1+0.04)

T=5.7 years

D) t=log(3,000÷2,000)÷log(1+0.04)

t=10.3 years

Hope it helps

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How would you write x+3y=3 in y=mx+b format

Triss [41]

X+3y=3
First add subtract the x
3y=-x+3
Then divide 3
Y=-1/3x+1

8 0

3 years ago

Read 2 more answers

What term is 1/1024 in the geometric sequence,-1,1/4,-1/6..?

Trava [24]

Answer:

$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

Step-by-step explanation:

The explicit formula for a geometric sequence:

$a_n=a_1r^{n-1}$

$a_n$ - n-th term

$a_1$ - first term

$r$ - common ratio

$r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}$

We have

$a_1=-1,\ a_2=\dfrac{1}{4},\ a_3=-\dfrac{1}{6},\ ...$

The common ratio:

$r=\dfrac{\frac{1}{4}}{-1}=-\dfrac{1}{4}\\\\r=\dfrac{-\frac{1}{6}}{\frac{1}{4}}=-\dfrac{1}{6}\cdot\dfrac{4}{1}=-\dfrac{2}{3}\neq-\dfrac{1}{4}$

<h2>It's not a geometric sequence.</h2>

If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

Put to the explicit formula:

$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

Put $a_n=\dfrac{1}{1024}$ and solve for <em>n </em>:

$-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1$

$\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6$

5 0

3 years ago

Stephanie opened a savings account with $200.00. Each additional month, the account will increase by 10%. The function f(x)=200

spin [16.1K]

<span>B.
approximately $325</span>

6 0

3 years ago

A random sample of 20 purchases showed the amounts in the table (in $). The mean is $49.57 and the standard deviation is $20.28.

son4ous [18]

Answer:

The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.325$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.325*\frac{20.28}{\sqrt{20}} = 12.17$

The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.

The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.

The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).

3 0

3 years ago