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Lisa [10]
3 years ago
9

Shannon is selling boxes cookies and candy for a fundraiser. She's sold each box of cookies for $5.25. She also sold one one box

of candies for $8.75.
Shannon set a goal to sell at least $200 of cookies and candies for her fundraiser.

What was the minimum boxes of cookies Shannon needed to sell to reach her goal?

Mathematics
1 answer:
defon3 years ago
5 0

Answer:

The minimum boxes of cookies is 39

Step-by-step explanation:

Let

x ----> the number of boxes of cookies sold

y ----> the number of boxes of candies sold

we know that

The word "at least" means "greater than or equal to"

so

The inequality that represent this problem is

5.25x+8.75y\geq 200

The solution is the shaded area above the solid line   5.25x+8.75y=200

using a graphing tool

The solution is the shaded area -----> see the attached figure

Find out the minimum boxes of cookies needed to sell to reach the goal

assuming only cookies are sold

For y=0

5.25x+8.75(0)\geq 200

5.25x\geq 200

solve for x

x\geq 38.1

Round up

therefore

The minimum boxes of cookies is 39

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An article reports on an experiment in which each of five groups consisting of six rats was put on a diet with a different carbo
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Answer:

H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

Step-by-step explanation:

<u> A              B               C          D              E</u>

<u>2.58 2.63 2.13          2.41    2.49</u>

∑A=2.58 ∑B=2.63 ∑C=2.13  ∑D=2.41   ∑E=2.49

<u>A²                           B²                C²               D²             E²</u>

<u>6.6564               6.9169 4.5369             5.8081 6.2001</u>

∑A²=6.6564 ∑B²=6.9169 ∑C²=4.5369 ∑D²=5.8081 ∑E²=6.2001

Calculations

<u>Group A            B           C           D             E       Total</u>

<u>N      n1=1         n2=1 n3=1 n4=1 n5=1 n=5</u>

∑xi T1=2.58    T2=2.63 T3=2.13 T4=2.41 T5=2.49 ∑x=12.24

<u>∑x²i  6.6564      6.9169 4.5369 5.8081 6.2001 ∑x²=30.1184</u>

<u>Mean xi 2.58 2.63 2.13          2.41         2.49 Total2.448</u>

Σ Σ xij^2 = 185.4 given

Let k = the number of different samples = 5

n=n1+n2+n3+n4+n5=1+1+1+1+1=5

Overall ˉx=12.24/5  =2.448

∑x=T1+T2+T3+T4+T5=2.58+2.63+2.13+2.41+2.49=12.24----1

Correction Factor=(∑x)²/n=12.242/5=29.9635--------(2)

∑T²i/ni=(2.582/1+2.632/1+2.132/1+2.412/1+2.492/1)=30.1184------(3)

∑x²=∑x²1+∑x²2+∑x²3+∑x²4+∑x²5

=6.6564+6.9169+4.5369+5.8081+6.2001=30.1184-------(4)

ANOVA:

Step-1 : sum of squares between samples

SSB=(∑T2i/ni)-(∑x)2/n

=(3)-(2)

=30.1184-29.9635

=0.1549

Step-2 : Total sum of squares

SST=SSB-CF

= 185.4-2.488

=182.952

Step-3 : Sum of Squares Within Samples

SSW=∑x²-(∑T²i/ni)=SST- SSB=

        = 182.952-0.1549=182.7971

Step-4 : variance between samples

MSB=SSB/k-1

=0.1549/4

=0.0387

Step-5 : variance within samples

MSW=SSW/n-k

=182.7971/5-5

=36.55942

Step-6 : test statistic F for one way ANOVA test

F=MSB/MSW

=0.0387/36.55942 = 0 .001058

<u>ANOVA table</u>

Source               Sums                  Degrees        Mean

of Variation    of Squares       of freedom        Squares

                                SS                        DF                 MS              F

B/w samples         SSB = 0.1549        k-1 = 4 MSB = 0.0387 0.001

<u>Within samples  SSW = 0               n-k = 0 MSW = 36.55942  </u>

<u>Total                  SST = 0.1549       n-1 = 4   </u>

H0 : The true average DNA content is not affected by the type of carbohydrate in the diet between samples

Ha : The true average DNA content is affected by the type of carbohydrate in the diet between samples

F(4,0) at 0.05 level of significance =0.5

As calculated F=0.001 >0.5

So, H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

6 0
3 years ago
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