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4 years ago

PLZ I REALLY NEED HELP!

Mathematics

1 answer:

allochka39001 [22]4 years ago

6 0

1.) That just means that X will equal 8, so the same as:
3(8)-1
24-1
23

2.) The relationship is a function, there are no repeated elements in the domain(x column), and each domain is pair with one element in the range.
The function is y = 2x+1, when you double x and add 1, you get y.

3.) The relation is not a function for reasons I explained in 2.), x is paired with more than 1 element. The domain is 1,3,-5. Range is -2,2,0,4,-4

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Chef Selena is making her special dish that requires pasta, sauce, and onions. She has 5 1/8 pounds of pasta, 6 pounds of sauce,

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Answer:

$\text{Total Weight of Food Made}, x=5\frac{1}{8}+6+4\frac{7}{8}\\\text{Number of Servings}, y=x \div \frac{1}{2}$

Step-by-step explanation:

The Weight of Ingredients to be used by Chef Selena are:

$5\frac{1}{8}$ pounds of pasta,
6 pounds of sauce; and
$4\frac{7}{8}$ pounds of onions.

Total Weight of Food Made, x

$=5\frac{1}{8}+6+4\frac{7}{8}\\=5+6+4+\frac{1}{8}+\frac{7}{8}\\=16 \: Pounds$

Since each serving is $\frac{1}{2} Pounds$

Number of Servings of her dish, y

$16 \div \frac{1}{2}\\ =32 \: Servings$

The two equations needed to solve the problem are therefore:

$\text{Total Weight of Food Made}, x=5\frac{1}{8}+6+4\frac{7}{8}\\\text{Number of Servings}, y=x \div \frac{1}{2}$

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3 years ago

Southwest Co-op produces bags of fertilizer, and it is concerned about impurity content. It is believed that the weights of impu

Harman [31]

Answer:

a) 0.216

b) 0.1587

c) 0.369

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12.2 grams

Standard Deviation, σ = 2.8 grams

We are given that the distribution of weights of impurities per bag is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(contains less than 10 grams of impurities)

P(x < 10)

$P( x < 10) = P( z < \displaystyle\frac{10 - 12.2}{2.8}) = P(z < -0.7857)$

Calculation the value from standard normal z table, we have,

$P(x < 10) =0.216= 21.6\%$

b) P(contains more than 15 grams of impurities)

P(x > 15)

$P( x > 15) = P( z > \displaystyle\frac{15 - 12.2}{2.8}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 610) = 1 - 0.8413 = 0.1587 = 15.87\%$

c) P(contains between 12 and 15 grams of impurities)

$P(12 \leq x \leq 15) = P(\displaystyle\frac{12 - 12.2}{2.8} \leq z \leq \displaystyle\frac{15-12.2}{2.8}) \\\\= P(-0.0714 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -0.0714)\\= 0.841 - 0.472 = 0.369 = 36.9\%$

$P(12 \leq x \leq 15) = 36.9%$

d) The mean divide the data in exactly two parts. Since 15 is farther away from the mean as compared to 10, the probability obtained in part (b) is smaller as compared to probability obtained in part (a).

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Answer:

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