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1 year ago

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You ask 150 people about their pets. The results show that 9/25 of the people own a dog. Of the people who own a dog. 1/6 of the

m also own a cat. How many people own a dog and a cat?

1 answer:

pav-90 [236]1 year ago

6 0

Answer: 9

Step-by-step explanation:

In order to find the people out of the 150, we first multiply 9/25 and 150 together. This will give us the people who DO own a dog. This brings us to 54.

Now, if we multiply 54 by 1/6 we get the answer of 9, because 1/6 of the dog owners own a cat as well. We have now reached the desired amount.

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Adjust point D so the measure of angle BAD is equal to the measure of angle CAD. Which statements are true? Check all that apply

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Answer with explanation:

it is given that ,in ΔABC,

∠BAD=20°, and ∠CAD=54°

We have to adjust point D,so that measure of angle BAD is equal to the measure of angle CAD.

that is, if point D is moved to right of B,then ∠BAD increases from 20° to (20+x)° and ∠CAD decreases from 54° to (54-x)°.

→20 +x=54-x

⇒ 2 x= 54 -20

⇒2 x=34

x=17°

Using angle bisector theorem, if AD bisects ∠B AC.

$\frac{BA}{AC}=\frac{BD}{DC}$

so,

If, ∠BAD=∠CAD=37°, then

1. AD bisects ∠B AC.→→→Option A

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A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

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Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$

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2 years ago

Solve the system of linear equations using multiplication.

Anna71 [15]

Answer:

$(8,-1)$

Step-by-step explanation:

The given system is:

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Let's do that.

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Subtract $y$ on both sides:

$x=7-y$

So we are replacing the second $x$ in the second equation with $(7-y)$ which gives us:

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$7-y+2y=6$

$7+y=6$

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$y=-1$

Now recall the first equation we arranged so that $x$ was the subject. I'm referring to $x=7-y$ .

We can now find $x$ given that $y=-1$ using the equation $x=7-y$ .

Let's do that.

$x=7-y$ with $y=-1$ :

$x=7-(-1)$

$x=7+1$

$x=8$

So the solution is (8,-1).

We can check this point by plugging it into both equations.

If both equations render true for that point, then we have verify the solution.

Let's try it.

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$3(8)+3(-1)=21$

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$21=21$ is a true equation so the "solution" looks promising still.

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$36=36$ is also true so the solution has been verified since both equations render true for that point.

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