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3 years ago

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Drivers who are members of the teamsters Union earn an average of $17.15 per hour (U.S. News & World Report). Assume that av

ailable data indicate wages are normally distributed with a standard deviation of $2.25. 1) What is the probability that wages are between $15.00 and $20.00 per hours?

1 answer:

Sladkaya [172]3 years ago

8 0

Answer:

$P(15$

And we can find the probability with this difference and using the normal standard table:

$P(-0.96$

Step-by-step explanation:

Let X the random variable that represent the wages, and for this case we know the distribution for X is given by:

$X \sim N(17.15,2.25)$

Where $\mu=17.15$ and $\sigma=2.25$

We want to find this probability:

$P(15$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

Using this formula we have:

$P(15$

And we can find the probability with this difference and using the normal standard table:

$P(-0.96$

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She bought a toy for 4 dollars, so we need to subtract that from 88. This gives us 84 dolalrs. She bought 8 shirts and 6 pants, which is 14 items total. Divide 88 by 14 and you get 6.

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Step-by-step explanation:

Lets call the amount of toppings that Ron got t. With this, we can set up the following equation:

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3 years ago

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Suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg,

lianna [129]

Answer:

(a) Probability that Y falls into the dangerous region is 0.0013.

(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001.

Step-by-step explanation:

We are given that a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg, σ = 1 mg).

A dosage of 13 mg is considered dangerous.

Let Y = <u><em>dosage of the active ingredient </em></u>

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ Y-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

(a) Probability that Y falls into the dangerous region is given by = P(Y $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{ Y-\mu}{\sigma} } }$ $\geq$ $\frac{ 13-10}{1} } }$ ) = P(Z $\geq$ 3) = 1 - P(Z < 3)

= 1 - 0.9987 = <u>0.0013</u>

The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.

(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.

Let $\bar Y$ = sample mean dosage

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ ~ N(0,1)

where, $\mu$ = population mean = 10 mg

$\sigma$ = standard deviation = 1 mg

n = sample of capsules = 49

So, Probability that the mean Y-bar falls into the dangerous region is given by = P( $\bar Y$ $\geq$ 13 mg)

P(Y $\geq$ 13 mg) = P( $\frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }$ $\geq$ $\frac{13-10}{\frac{1}{\sqrt{49} } } } }$ ) = P(Z $\geq$ 21) = 1 - P(Z < 21)

= <u>0.00001</u>

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3 years ago

One factor of the function f(x) = x3 − 8x2 + 17x − 10 is (x − 5). Describe how to find the x-intercepts and the y-intercept of t

kozerog [31]

Answer:

x-intercepts: (1,0), (2,0), (5,0)

y-intercepts: (0,-10)

Step-by-step explanation:

One factor of the function $f(x) = x^3-8x^2 + 17x-10$ is $(x-5).$

Factor the polynomial in the following way:

$x^3-8x^2 + 17x-10\\ \\=x^3-5x^2-3x^2+15x+2x-10\\ \\=x^2(x-5)-3x(x-5)+2(x-5)\\ \\=(x-5)(x^2-3x+2)$

Now factor the quadratic in brackets:

$x^2-3x+2\\ \\=x^2-2x-x+2\\ \\=x(x-2)-(x-2)\\ \\=(x-2)(x-1)$

Hence,

$f(x)=(x-5)(x-2)(x-1)$

The x-intercepts are points at which y = 0, so

$(x-5)(x-2)(x-1)=0\\ \\x-5=0\ \text{or}\ x-2=0\ \text{or}\ x-1=0\\ \\x=5\ \text{or}\ x=2\ \text{or}\ x=1$

The y-intercepts are points at which x = 0, so

$y=0^3-8\cdot 0^2+17\cdot 0-10=-10$

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