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olga_2 [115]
1 year ago
13

A recipe calls for 2 tsp of salt, 1 tsp of pepper, and 4 tsp of garlic powder. How much pepper and garlic powder would you need

for a larger batch if you are using 15 tsp of salt?
Mathematics
1 answer:
MakcuM [25]1 year ago
3 0

Answer:

amount of pepper required= 7.5 tsp

amount of garlic powder required = 30 tsp

Step-by-step explanation:

Given,

amount of salt used for small batch of the recipe = 2 tsp

amount of pepper used for small batch of the recipe = 1 tsp

amount of garlic powder used for small batch of the recipe = 4 tsp

amount of salt used for the larger batch = 15 tsp

                                                             = 2 x 7.5 tsp

                                                             = amount of salt used for small batch the recipe x 7.5

So,

the amount of pepper needed for the larger batch= 7.5 x amount of pepper used for the small batch of recipe

                                                                      = 7.5 x 1 tsp

                                                                       = 7.5 tsp

the amount of garlic powder needed for the larger batch= 7.5 x amount of garlic powder used for the small batch of recipe

                                                                                          = 7.5 x 4 tsp

                                                                                          = 30 tsp

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Step-by-step explanation:

If the total games played was 36 and no team played each other twice, we need to ensure there isn't any double counting.

36 = (n-1) + (n-2) + (n-3) ... + (n-(n-1))

using this knowledge, we can then count up:

1+2+3+4+5+6+7+8 = 36

If our highest number is 8, then we know there must be 9 teams, because no team can play themselves.

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Answer:

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In order to find the variance we need to find first the second moment given by:

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And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

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LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

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P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

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calculeaza lungimea segmentului ab in fiecare dintre cazuri:A(1,5);B(4,5);A(2,-5),B(2,7);A(3,1)B(-1,4);A(-2,-5)B(3,7);A(5,4);B(-
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Step-by-step explanation:

We can use the distance formula to calculate the lengths of the line segments.

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1. A (1,5), B (4,5) (red)

d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}

2. A (2,-5), B (2,7) (blue)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}

3. A (3,1), B (-1,4 ) (green)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}

4. A (-2,-5), B (3,7) (orange)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(3 - (-2))^{2} + (7 - (-5))^{2}}\\= \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}

5. A (5,4), B (-3,-2) (purple)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-3 - 5)^{2} + (-2 - 4)^{2}}\\= \sqrt{(-8)^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}

6. A (1,-8), B (-5,0) (black)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-5 - 1)^{2} + (0 - (-8))^{2}}\\-= \sqrt{(-6)^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}

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