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2 years ago

A) .25 b) .30 c) .45 d) .50

Mathematics

1 answer:

kupik [55]2 years ago

3 0

Answer:

0.45

Step-by-step explanation:

Since the percent to pull a ball with a 2 is 30% and the percent to pull a ball with a 4 is 15%, that means that the chance of pulling a ball with an even number is 0.45, or 45%. (0.30+0.15=0.45)

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(1 point) Suppose F⃗ (x,y)=⟨2y,−sin(y)⟩ and C is the circle of radius 3 centered at the origin oriented counterclockwise. (a) Fi

g100num [7]

Answer:

The required vector parametric equation is given as:

r(t) = <3cost, 3sint>

For 0 ≤ t ≤ 2π

Step-by-step explanation:

Given that

f(x, y) = <2y, -sin(y)>

Since C is a cirlce centered at the origin (0, 0), with radius r = 3, it takes the form

(x - 0)² + (y - 0)² = r²

Which is

x² + y² = 9

Because

cos²β + sin²β = 1

and we want to find a vector parametric equations r(t) for the circle C that starts at the point (3, 0), we can write

x = 3cosβ

y = 3sinβ

So that

x² + y² = 3²cos²β + 3²sin²β

= 9(cos²β + sin²β) = 9

That is

x² + y² = 9

The vector parametric equation r(t) is therefore given as

r(t) = <x(t), y(t)>

= <3cost, 3sint>

For 0 ≤ t ≤ 2π

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3 years ago

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36.91 is the answer 67.1x.55

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An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$