If you triple the amount of ground ginger to make more pies, how much nutmeg should you use to keep the recipe proportional

If a unit price label has a unit price of 13.5 cents per ounce, how much is the total price for a 24 ounce package? a. $0.32 b.

marta [7]

Multiply:
13.5*24=324cents=$3.24

8 0

3 years ago

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For ten school days the numbers of bikes parked at a school bike rack are 10,12,8,11,13,9,2,1,9, and 12

kumpel [21]

0

Step-by-step explanation:

for this question is not even possible

6 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20-%205%282y%20-%208%29%20-%203%281%20%2B%20x%29%20-%207" id="TexFormula1" title=" - 5(2y - 8

Andre45 [30]

-(2y-8)-3(1+x)-7
-10y+40-3(1-x)-7
-10y+40-3-3x-7
-10y+30-3x

5 0

3 years ago

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How is the series 5+11+17…+251 represented in summation notation?

NISA [10]

Answer:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

Step-by-step explanation:

Given:

Series 5+11+17+...+251

To find:

Summation notation of the given series

Summation Notation:

$\displaystyle \large{\sum_{k=1}^n a_k}$

Where n is the number of terms and $\displaystyle \large{a_k}$ is general term.

First, determine what kind of series it is, there are two main series that everyone should know:

Arithmetic Series

A series that has common difference.

Geometric Series

A series that has common ratio.

If you notice and keep subtracting the next term with previous term:

11-5 = 6
17-11 = 6

Two common difference, we can in fact say that the series is arithmetic one. Since we know the type of series, we have to find the number of terms.

Now that brings us to arithmetic sequence, we know that first term is 5 and last term is 251, we’ll be finding both general term and number of term using arithmetic sequence:

<u>Arithmetic Sequence</u>

$\displaystyle \large{a_n=a_1+(n-1)d}$

Where $\displaystyle \large{a_n}$ is the nth term, $\displaystyle \large{a_1}$ is the first term and $\displaystyle \large{d}$ is the common difference:

So for our general term:

$\displaystyle \large{a_n=5+(n-1)6}\\\displaystyle \large{a_n=5+6n-6}\\\displaystyle \large{a_n=6n-1}$

And for number of terms, substitute $\displaystyle \large{a_n}$ = 251 and solve for n:

$\displaystyle \large{251=6n-1}\\\displaystyle \large{252=6n}\\\displaystyle \large{n=42}$

Now we can convert the series to summation notation as given the formula above, substitute as we get:

$\displaystyle \large{\sum_{n=1}^{42}(6n-1)$

5 0

2 years ago