The graph represents the normal distribution of recorded weights, in pounds, of cats at a veterinary clinic.

Mathematics

2 answers:

diamong [38]3 years ago

3 0

Answer:

B) 8.9 lbs

C) 9.5 lbs

D) 9.8 lbs

E) 10.4 lbs

Step-by-step explanation:

From the graph, 9.5 is the mean of the sample and 0.5 is the standard deviation of the sample.

As we have to find the weights that lie within the 2 standard deviations of the mean i.e

$=9.5\pm 2(0.5)$

$=8.5,10.5$

Among the given weights only 8.9 lbs, 9.5 lbs, 9.8 lbs, 10.4 lbs will lie within 2 standard deviations of the mean.

anyanavicka [17]3 years ago

3 0

Answer:

Step-by-step explanation:

8.9

9.5

9.8

10.4

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Admission to the zoo is $6 for children and $9 for adults. On a certain day, 1480 people go to the zoo and $10,500 is collected.

erik [133]

6x+9y = 10500
x+y = 1480

Using elimination I get: 6x + 9y = 10500 -6x-6y = -8880
Add the 2 equations and I get 3y = 1620
Divide both sides by 3 and I get y = 540
Substitute that back in to 6x+9y = 10500.
I get that 6x+9*540 = 10500 9 * 540 = 4860
Subtract 4860 from both sides and I get 6x = 5640
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The ability of lizards to recognize their predators via tongue flicks can often mean life or death for lizards. Seventeen juveni

Sloan [31]

Answer:

Confidence level is (380.7133, 533.2867)

Step-by-step explanation:

Responses in number of tongue flicks per 20 minutes of lizards, are: 727,217, 268, 438, 625, 319, 200, 591, 574, 727, 693, 336, 302, 761, 268, 353, 370

n = 17

Mean (μ) is:

$\mu=\frac{{\Sigma}x}{n} = \frac{727+217+ 268+ 438+ 625+ 319+ 200+ 591+ 574+ 727+ 693+ 336+ 302+ 761+ 268+ 353+ 370}{17} = 439.3529$