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2 years ago

11

A cup is filled with pencils. The teacher sharpen 1/4 of them. A student sharpen 2/3 of them. The fraction of pencils left to be

sharpened is?

1 answer:

Bad White [126]2 years ago

4 0

Answer:

1/12

Step-by-step explanation:

1/4+2/3=11/12

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A company is producing two types of ski goggles. Thirty percent of the production is of type A, and the rest is of type B. Five

exis [7]

Answer:

$\dfrac{14}{29}$

Step-by-step explanation:

Let <em>P(A) </em>be the probability that goggle of type A is manufactured

<em>P(B) </em>be the probability that goggle of type B is manufactured

<em>P(E)</em> be the probability that a goggle is returned within 10 days of its purchase.

According to the question,

<em>P(A)</em> = 30%

<em>P(B)</em> = 70%

<em>P(E/A)</em> is the probability that a goggle is returned within 10 days of its purchase given that it was of type A.

P(E/B) is the probability that a goggle is returned within 10 days of its purchase given that it was of type B.

$P(A \cap E)$ will be the probability that a goggle is of type A and is returned within 10 days of its purchase.

$P(B \cap E)$ will be the probability that a goggle is of type B and is returned within 10 days of its purchase.

$P(E \cap A) = P(A) \times P(E/A)$

$P(E \cap A) = \dfrac{30}{100} \times \dfrac{5}{100}\\\Rightarrow P(E \cap A) = 1.5 \%$

$P(E \cap B) = P(B) \times P(E/B)$

$P(E \cap B) = \dfrac{70}{100} \times \dfrac{2}{100}\\\Rightarrow P(E \cap B) = 1.4 \%$

$P(E) = 1.5 \% + 1.4 \% \\P(E) = 2.9\%$

If a goggle is returned within 10 days of its purchase, probability that it was of type B:

$P(B/E) = \dfrac{P(E \cap B)}{P(E)}$

$\Rightarrow \dfrac{1.4 \%}{2.9\%}\\\Rightarrow \dfrac{14}{29}$

So, the required probability is $\dfrac{14}{29}.$

7 0

2 years ago

Tamiku [17]

Answer:

<h2> x₁ = - 2 + √2 , x₂ = - 2 - √2</h2>

Step-by-step explanation:

$3x^2 + 12x + 6 = 0\\\\a=3\,,\ b=12\,,\ c=6\\\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-12\pm\sqrt{12^2-4\cdot3\cdot6}}{2\cdot3}=\dfrac{-12\pm\sqrt{144-72}}{2\cdot3}=\dfrac{-12\pm\sqrt{72}}{2\cdot3}=\dfrac{-12\pm6\sqrt{2}}{6}\\\\x_1=\dfrac{6(-2+\sqrt{2})}{6}=-2+\sqrt2\\\\x_2=\dfrac{6(-2-\sqrt{2})}{6}=-2-\sqrt2$

8 0

2 years ago

Plsss helpppp what’s the shape?

Alborosie

The shape shown here is a triangular pyramid, because it has a triangular base and a pointy tip.

4 0

2 years ago

Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig

Anon25 [30]

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3466, \sigma = 546$

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2500 - 3466}{546}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384

3.84% probability that it has a low birth weight

3 0

2 years ago

Um i need help with this.

blsea [12.9K]

Answer:

None of these!

Step-by-step explanation:

x + y + z = x + y + z only over here

5 0

2 years ago

Read 2 more answers