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choli [55]
2 years ago
7

According to a report for veterinarians in the United States, 36.5 percent of households in the United States own dogs and 30.4

percent of households in the United States own cats. If one household in the United States is selected at random, what is the probability that the selected household will own a dog or a cat?(A)0.111(B) 0.331(C) 0.558(D) 0.669(E) Not enough information is given to determine the probability.
Mathematics
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer :E) Not enough information is given to determine the probability.

Step-by-step explanation:

Le A denotes the event that households in the United States own dogs .

and B denotes the event that households in the United States own cats.

As per given , we have

P(A)=36.5%= 0.365

P(B)=30.4% = 0.304

To find the probability that the selected household will own a dog or a cat, we apply the following formula :

P(A or B)=P(A)+P(B)+P(A and B)

But P(A and B) is not given to us.

i..e the probability that a house hold own both a cat and adg is not given to us.

Therefore, The correct option is (E) Not enough information is given to determine the probability.

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Jordan had the following scores on her math tests last quarter: 96, 89, 79, 85, 87, 94,
Lisa [10]

Answer:

a) 5.5

b) None

Step-by-step explanation:

The given data set is {96,89,79,85,87,94,96,98}

First we must find the mean.

\bar X=\frac{96+89+79+85+87+94+96+98}{8}=\frac{724}{8}=90.5

We now find the absolute value of the distance of each value from the mean.

This is called the absolute deviation

{|96-90.5|,|89-90.5|,|79-90.5|,|85-90.5|,|87-90.5|,|94-90.5|,|96-90.5|,|98-90.5|}

{5.5,1.5,11.5,5.5,3.5,3.5,5.5,7.5}

We now find the mean of the absolute deviations

MAD=\frac{5.5+1.5+11.5+5.5+3.5+3.5+5.5+7.5}{8} =\frac{44}{8} =5.5

The least absolute deviation is 1.5. This is not within one absolute deviation.

Therefore none of the data set is closer than one mean absolute deviation away from  the mean.

5 0
2 years ago
Choose the function whose graph is given by:
loris [4]
<h2>Answer:</h2>

y=3sec\left(\frac{1}{2}x\right)

<h2>Step-by-step explanation:</h2>

The graphs of sec(x) can be obtained from  the graph of the cosine function using the reciprocal identity, so:

sec(x)=\frac{1}{cos(x)}

But in this problem, the graph stands for the function:

y=3sec\left(\frac{1}{2}x\right)

Because the period is now 4π as indicated and for x=0 in the figure and this can be proven as follows:

Period=\frac{2\pi}{\frac{1}{2}}=4\pi

Also, for \ x=1 \ then \ y=3 as indicated in the figure and this can be proven as:

y=3sec\left(\frac{1}{2}x\right) \\ \\ y=\frac{3}{cos(0.5x)} \\ \\ y=\frac{3}{cos(0.5(0))} \\ \\ y=\frac{3}{1}=3

5 0
2 years ago
a) As a student, you are able to earn extra money by assisting your neighbors with odd jobs. If you charged $10.25 an hour for y
inna [77]
ABout 821.95 hours. 8,425 divided by 10.25 is 821.95

7 0
2 years ago
It takes 24 electricians 36 days to wire a new housing subdivision. How many days would 32 electricians take to do the same job?
Naya [18.7K]

Answer: 27 days

Step-by-step explanation:

Hi, to answer this question we have to apply inverse relation:

Since it takes 24 electricians 36 days to wire a new housing subdivision, the relation is:

24 (36)  

For 32 electricians:

24(36) = 32 (x)

Solving for x (days)

864 =32x

864/32 =x

x= 27 days

Feel free to ask for more if needed or if you did not understand something.

6 0
2 years ago
in the drawing six out of every 10 tickets are winning tickets of the winning tickets one out of every three awards is a larger
Dafna11 [192]

Answer: P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

Since we have given that

Total number of tickets =10

Number of winning tickets =6

\text {Possibility of getting a winning ticket }=^{10}C_6

In which one out of every three awards is a large prize

so,

\text{ Possibility of getting a large prize }=^6C_3

Now,

P(\text{a ticket is randomly chosen will award a large prize)}=\frac{^6C_3}{^{10}C_6}\\\\=\frac{20}{210}\\\\=\frac{2}{21}

P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

8 0
2 years ago
Read 2 more answers
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