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3 years ago

Log 3x = 4how do you cancel out the log?

1 answer:

8 0

This can be rewritten as log (with a base of 10) of 3x equals 4. The base, in this case 10, can be moved to the other side.
log3x=4 becomes 3x=10^4, and the log is gone.

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Given that a randomly chosen quadrilateral has four right

motikmotik

Answer:

0.25

Step-by-step explanation:

complete question:

Bart found 20 quadrilaterals in his classroom. He made a Venn diagram using the properties of the quadrilaterals, comparing those with four equal side lengths (E) and those with four right angles (R).

See attachment for the figure.

SOLUTION:

At Venn diagram there are 4 parts (20 pieces):

-> blue colored - quadrilaterals having four equal side lengths (3 pieces)

-> orange colored - quadrilaterals with four right angles (6 pieces)

-> blue and orange colored - quadrilaterals with four right angles and with four equal side lengths (2 pieces)

-> white colored - quadrilaterals without previous two properties (9 pieces).

Considering events:

A -> a randomly chosen quadrilateral has four right angles;

B -> a randomly chosen quadrilateral has four equal side lengths;

By using formula : $P(B|A)=\dfrac{Pr(A\cap B)}{Pr A}$ in order to find probability that a randomly selected quadrilateral with 4 right angles also has four equal side lengths:

$P(A\cap B)=\dfrac{2}{20},\\P(A)=\dfrac{8}{20},\\P(B|A)=\dfrac{\frac{2}{20}}{\frac{8}{20}} =\dfrac{2}{8}=\dfrac{1}{4} =0.25$

3 0

3 years ago

In studies for a medication, 3 percent of patients gained weight as a side effect. Suppose 643 patients are randomly selected.

timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

$p = 0.03$

$q = 1 - 0.03 = 0.97$

The mean is

$\mu = np$

$\mu = 643 \times 0.03 = 19.29$

The standard deviation is

$\sigma = \sqrt{npq}$

$\sigma = \sqrt{643 \times 0.03 \times 0.97}$

$\sigma =4.33$

We want to find the probability that exactly 24 patients will gain weight as side effect.

P(X=24)

We apply the Continuity Correction Factor(CCF)

P(24-0.5<X<24+0.5)=P(23.5<X<24.5)

We convert to z-scores.

$P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 1.20) \\ = 0.051$

Part b) We want to find the probability that 24 or fewer patients will gain weight as a side effect.

P(X≤24)

We apply the continuity correction factor to get;

P(X<24+0.5)=P(X<24.5)

We convert to z-scores to get:

$P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849$

Part c)

We want to find the probability that

11 or more patients will gain weight as a side effect.

P(X≥11)

Apply correction factor to get:

P(X>11-0.5)=P(X>10.5)

We convert to z-scores:

$P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)$

$= 0.9788$

Part d)

We want to find the probability that:

between 24 and 28, inclusive, will gain weight as a side effect.

P(24≤X≤28)=

P(23.5≤X≤28.5)

Convert to z-scores:

$P(23.5 \: < \: X \: < \: 28.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{28.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 2.13) \\ = 0.1494$