60 is 30% of what number?

1 answer:

7 0

Percent means parts out of 100
30%=30/100=3/10

'of' means multiply

60 is 30% of what translates to
60=3/10 times what
multiply both sides by 10/3
600/3=what
200=what
the number is 200

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It costs $12 to attend a golf clinic with a local pro. Buckets of balls for practice during the clinic cost $3 each. How many bu

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You can buy 6 buckets

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3 years ago

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Find the range of each function for the domain {-4, -2, 0, 1.5, 4}. f(x) = 5x^2 + 4

andrew11 [14]

Substitute x with the members of the domain.
f(x) = 5x² + 4

Substitute with the domain of -4
f(x) = 5x² + 4
f(-4) = 5(-4)² + 4
f(-4) = 5(16) + 4
f(-4) = 80 + 4
f(-4) = 84

Substitute with the domain of -2
f(x) = 5x² + 4
f(-2) = 5(-2)² + 4
f(-2) = 5(4) + 4
f(-2) = 20 + 4
f(-2) = 24

Substitute with the domain of 0
f(x) = 5x² + 4
f(0) = 5(0)² + 4
f(0) = 5(0) + 4
f(0) = 0 + 4
f(0) = 4

Substitute with the domain of 1.5
f(x) = 5x² + 4
f(1.5) = 5(1.5)² + 4
f(1.5) = 5(2.25) + 4
f(1.5) = 11.25 + 4
f(1.5) = 15.25

Substitute with the domain of 4
f(x) = 5x² + 4
f(4) = 5(4)² + 4
f(4) = 5(16) + 4
f(4) = 80 + 4
f(4) = 84

The range of the function for those domain is {4, 24, 15.25, 84}

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3 years ago

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 = dosage of the active ingredient

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 = 0.0013

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)