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lidiya [134]
1 year ago
6

The Sydney Morning Herald - February 8, 2004 reported that in 1961, the average number of children born to Australian women (3.5

5) was at its highest level since reliable records began in the 1920s. With only the fact of the mean being 3.55 children, can one make the claim that women were statistically more likely to have four children than any other number of children?
a. Agree, 4 children is the most typical number of children.
b. Agree, 3.55 is closer to 4 than any other whole number.
c. Disagree, the most typical number of children is 3.
d. Disagree, it is impossible to tell
Mathematics
1 answer:
Alborosie1 year ago
3 0

Answer:

<h2>A. Agree, 4 children is the most typical number of children.</h2>

Step-by-step explanation:

By analysis the problem, we can make our  decision by first determining the percent of 3.55 of 4

therefore the percentage can be computed as

=(3.55/4)*100

=0.8875*100

=88.75%

The analysis above it shows that 88.75 percent of the women who give birth to children give birth to 4 children.

This number is very close to a hundred percent hence the statistical claim is  "Agree, 4 children is the most typical number of children."

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Answer:

35

Step-by-step explanation:

Here we see 5 black keys for every 7 white keys.

So the ratio is 5:7

If we need 49 white keys, find the amount we scale the original ratio by:

    49/7 = 7

So we are scaling by a factor of 7.

The number of black keys would be 5 * the scale of 7. = 35

So there should be 35 black keys.

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1 year ago
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Bailey likes the shape of her friend's pool, but she wants one that is half the volume. According to Cavalieri’s Principle, what
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Answer: I need a picture off the pool and measurements.

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2 years ago
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In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

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Afina-wow [57]
Ho ho ho, lets get this party started
ok so I'm just really excited to use this stuff that I just learned

so

multiplicites
if a root or zero has an even multilicity, the graph bounces on that root
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so
roots are
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2
4
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for roots, r1 and r2, the facotrs would be
(x-r1)(x-r2)
so
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(x+1)(x-2)^2(x-4)
this is a 4th degre equaton
normally, it is goig from top right to top left
it is upside down
theefor it has negative leading coefient



y=-k(x+1)(x-4)(x-2)^2
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