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3 years ago

20. A rectangle has an area of 45 cm². The length is one cm less than twice the width. Find the dimensions of

1 answer:

7 0

Width = 5, length = 9

Area = length * width

length = (2 * width) - 1

sub equation for length

45 = (2w - 1) * w
45 = 2w^2 - w
2w^2 - w - 45 = 0
(2w - 9)(w - 5)
2w - 9 = 0
w = 9/2, l = 8 no

w -5 = 0
w = 5, l = 9 yes

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Find the indicated limit, if it exists. limit of f of x as x approaches 9 where f of x equals x plus 9 when x is less than 9 and

SVEN [57.7K]

Answer:

B. 18

Step-by-step explanation:

For the function

$f(x)=\left\{\begin{array}{l}x+9,\ \ x$

we can find the value of the function for all x that are very close to 9 but are less than 9 and for all values of x that are very close to 9 but are greater than 9.

1. For $x$

$\lim \limits_{x\rightarrow 9}f(x)=\lim \limits_{x\rightarrow 9}(x+9)=9+9=18$

2. For $x\ge 9:$

$\lim \limits_{x\rightarrow 9}f(x)=\lim \limits_{x\rightarrow 9}(27-x)=27-9=18$

So, limit exists and is equal to 18.

8 0

3 years ago

According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what i

Ierofanga [76]

Answer:

a) $P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

b) $P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)$

$P(X=0) = (10C0) (0.2)^0 (1-0.2)^{10-0}= 0.107$

$P(X=1) = (10C1) (0.2)^1 (1-0.2)^{10-1}= 0.268$

$P(X=2) = (10C2) (0.2)^2 (1-0.2)^{10-2}= 0.302$

And replacing we got:

$P(X\leq 2) = 0.107+0.268+0.302=0.678$

c) For this case we want this probability:

$P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)$

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

$P(X=8) = (10C8) (0.8)^8 (1-0.8)^{10-8} =0.302$

$P(X=9) = (10C9) (0.8)^9 (1-0.8)^{10-9} =0.268$

$P(X=10) = (10C10) (0.8)^{10} (1-0.8)^{10-10} =0.107$

And replacing we got:

$P(X \geq 8) = 0.677$

And replacing we got:

$P(X\geq 8)=0.0000779$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Bin (n=10 ,p=0.2)$