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Maurinko [17]
3 years ago
14

1/5 of a chocolate chip cookie has 30 cal how many calories are in a whole cookie

Mathematics
2 answers:
krok68 [10]3 years ago
8 0

Answer:

150 cal

Step-by-step explanation:

5x30=150

Svetach [21]3 years ago
6 0

Answer:

150 calories.

Step-by-step explanation:

Assuming there is the same amount of chocolate as well as cookie dough throughout the whole cookie.

You know that 1/5 of a chocolate chip cookie has 30 calories.

Find one cookie, by multiply 5 to both numbers. Set the equation:

1/5x = 30

Isolate the variable. Multiply 5 to both sides:

(1/5x) * 5 = (30) * 5

x = 30 * 5

x = 150

150 calories is your answer.

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The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

Now Dbar = Bbar - Abar = 489 - 514 = -25

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

                            = \frac{-25 - 0}{\frac{37.05}{\sqrt{12} } } follows t_1_1   = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0
3 years ago
Substitute x=y+1 into the expression 3x -1 and determine the first step
Kruka [31]

Answer: the solution becomes 3=y+1-1

Step-by-step explanation:

The first step is to combine like terms meaning you subtract 1-1. Then you are left with 3=y.

8 0
3 years ago
Pedro fills a glass 2/4 full with orange juice. Write a fraction with a denominator of 6 that is greater than 2/4.
Mice21 [21]
The answer is 3/6.
So we know that 2/4 is equivalent to 1/2. 1/2 of 6 is 3. We can make this our numerator and our denominator as 6.
4 0
3 years ago
-9 +m=20<br> answer asap please!!
grigory [225]

Answer:

29

Step-by-step explanation:

at 9 on both sides to get m alone

-9+m=20

+9. +9

m= 29

7 0
2 years ago
Read 2 more answers
) A coin collection consisting of pennies, dimes, and quarters totals 45 coins. The number of
Sloan [31]

9514 1404 393

Answer:

  $7.14

Step-by-step explanation:

Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.

  p + d + q = 45 . . . . . . . . there are 45 coins in the collection

  2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies

  p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies

Substituting the last two equations into the first gives ...

  p +(p +4) +(2p +5) = 45

  4p = 36 . . . . . . . . . . . . . subtract 9

  p = 9 . . . . . . . . . . . divide by 4

  d = 9 +4 = 13

  q = 2(9) +5 = 23

The value of the collection is ...

  23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14

The coin collection is worth $7.14.

5 0
3 years ago
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