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Tom [10]
3 years ago
15

Many baseball pitchers can throw 3 different types of pitches: fastball, curve, and change-up. When facing a batter with only on

e strike left, a
certain pitcher throws his fastball 45% of the time, curveball 35% of the time, and change up the remaining times. The pitcher gets the batter
out 75% of the time when he throws a fastball, 80% of the time when he throws a curveball, and 65% of the time when he throws a change-up.
Randomly select one batter during a game.
P(Fastball and out) = ?
Mathematics
1 answer:
OLEGan [10]3 years ago
5 0

Answer:

125

Step-by-step explanation:

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Based on the graph, in which two months
Rudiy27

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Where is the graph?

Step-by-step explanation:

5 0
2 years ago
Please help me please will give brainliest to ​
KiRa [710]

Answer:

Step-by-step explanation:

No because √2 and 3i come in pairs. The four roots are ±√2, 3i and -3i. The 5 guarantees that there are at least 5 roots to this equation. This is a good question to know the answer to. It looks like something that could be put on a test.

3 0
3 years ago
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
3 years ago
A woman wishes to rent a house within 9 miles of her work. If she lives x miles from her work, her transportation cost will be c
I am Lyosha [343]

Answer:

the woman has to live 1 mile from work to minimize the expenses

Step-by-step explanation:

Given the data in the question;

the distance within 9 miles ⇒ 0 < x > 9

Total costs Q = cx + 4c/( x + 1)

costs should be minimum  ⇒ dQ/dx = 0

⇒ d/dx [ cx + 4c/( x + 1) ] = 0

⇒ ( x + 1)² = 4

take square root of both side

√[ ( x + 1)² ] = √4

x + 1 = 2

x = 2 - 1

x = 1

Therefore, the woman has to live 1 mile from work to minimize the expenses

5 0
3 years ago
On the box and whisker plot below, what data calculations do the points B and D represent?
daser333 [38]

Answer:

Lower and upper quartiles

Step-by-step explanation:

The box starts at lower quartile, ends at upper quartile

5 0
3 years ago
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