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2 years ago

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A recipe for whole grain bread calls for 1 oz of rye flour for every 3 oz of whole wheat flour How many ounces of each kind of f

lour should be used to make a loaf of
bread weighing 22 oz?

1 answer:

natali 33 [55]2 years ago

3 0

Given :

A recipe for whole grain bread calls for 1 oz of rye flour for every 3 oz of whole wheat flour.

To Find :

How many ounces of each kind of flour should be used to make a loaf of

bread weighing 22 oz.

Solution :

It is given that ratio of rye and whole wheat flour is 1 : 3.

Let, amount of rye flour is x and amount of whole wheat flour is 3x.

So, x + 3x = 22

4x = 22

x = 5.5 oz

3x = 16.5 oz

Therefore, amount of rye flour and whole wheat flour is 5.5 and 16.5 oz respectively.

Hence, this is the required solution.

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lubasha [3.4K]

Answer: 4 hours

Step-by-step explanation:

okay so you have to know your times tables 2*4=8 so 20*40=80 then if he paid a tip of 10$ that would equal $90 witch is what he paid.

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2 years ago

Need math help. Thanks in advance.

Mademuasel [1]

9. D because it fits the description when it is graphed.
11. D because it has the correct vertex, which is (1.5, 4.75).
12. f(x)=-4(x-2)²+16 works with the table and provided graph!
13. 0 isn't a zero for f(x), so B is correct (the zeros are 1/2 and 4).
14. The width is about 3.831.
15. The answer is B because the equation for a parabola in this form is:
f(x) = (x - r₁)(x - r₂)

Every one of these except for 15 was solved using desmos, a graphing calculator.

Hope this helps!

7 0

2 years ago

14. The cost of 10 dozen of caps is Rs.800. How many caps can bebought for Rs.200.

Mashcka [7]

Answer:

Step-by-step explanation:

10 dozen is equal to 120 so ratio is 800/120 or 80/12

same ration on the other side but 200/x.Cross multiply and u get 3

7 0

3 years ago

The ordered pairs (1,3), (2,9), (3,27), (4,81), and (5,243) represent a function. What is a rule that represents this function?

weqwewe [10]

The rule is 3^x.

Check: 3^1 = 3; 3^2 = 9; 3^3 = 27 and 3^4 = 81

8 0

3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

<u>Lower half:</u>

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

<u>Upper half</u>

<u></u> $\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$ <u></u>

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

<u></u> $\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$ <u></u>

<u></u>

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

<u>Lower half:</u>

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

$Median = \frac{N + 1}{2}th$

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

$Median = \frac{1}{2}(7th + 8th)$

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question.

$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

Collect like terms

$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

$26= \frac{ 2nd + 12th+304}{14}$

Multiply through by 14

$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

From the updated box,

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We know that:

<em>The 2nd value can only be either 2 or 3</em>

<em>The 12th value can take any of the range 33 to 57</em>

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

$2nd = 3$

$12th = 57$

i.e.

$2nd + 12th = 60$

$3 + 57 = 60$

So, the complete box is:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

6 0

2 years ago