2 years ago

I need help plz plz plz

Mathematics

1 answer:

yan [13]2 years ago

8 0

Step-by-step explanation:

The answer is

X=23

Hope it helpsss

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C and D are mutually exclusive events. Find P(C or D).<br> P(C)= 3/7= P(D)= 4/7<br> P(C or D)

IgorLugansk [536]

Answer:

$P(C\ or\ D) = 1$

Step-by-step explanation:

Given

$P(C) = \frac{3}{7}$

$P(D) = \frac{4}{7}$

Required

$P(C\ or\ D)$

Since the events are mutually exclusive, then:

$P(C\ or\ D) = P(C) + P(D)$

So, we have:

$P(C\ or\ D) = \frac{3}{7} + \frac{4}{7}$

Take LCM

$P(C\ or\ D) = \frac{3+4}{7}$

$P(C\ or\ D) = \frac{7}{7}$

$P(C\ or\ D) = 1$

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Flauer [41]

Answer:

c=2(3.14)r

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c=2•3.14•5

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2 years ago

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dangina [55]

5 to the fifth power or 3125

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2 years ago

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Make a histogram, using a bin width of ten, to display the bowling scores for these 31 players: 87, 104, 79, 94, 117, 82, 72, 11

slavikrds [6]

Step-by-step explanation:

Using R, I used the following code to create a histogram:

bowling.scores <- c(87, 104, 79, 94, 117, 82, 72, 116, 105, 95, 88, 93, 109, 119, 75,

103, 112, 97, 73, 85, 91, 86, 102, 99, 106, 84, 98, 83, 81, 96, 92)

data.frame(bowling.scores)

ggplot(data.frame(bowling.scores), aes(x=bowling.scores)) +

xlim(c(70, 120)) +

scale_y_continuous(breaks = seq(0, 10, by=1), "Frequency") +

geom_histogram(breaks=seq(70, 120, by=10), color="black", fill="grey60") +

labs(title="Histogram of Bowling Scores", x="Bowling Scores", y="Frequency")

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3 years ago

Is number 4 correct?

Strike441 [17]

Answer:

sure

Step-by-step explanation:

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3 years ago