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3 years ago

20

Mathematics

1 answer:

Alexus [3.1K]3 years ago

3 0

Answer:

A wildlife biologist tracks the number of wolves in a particular region. Data for 15 months is shown. Use the

data to make a frequency table with intervals.

15, 19, 24, 28, 26, 29, 17, 16, 17, 14, 22, 20, 18, 27, 27

a. Wolves Observed

Number of Wolves Frequency

15  19 6

20  24 3

25  29 5

c. Wolves Observed

Number of Wolves Frequency

14  17 5

18  21 3

22  25 2

26  29 5

b. Wolves Observed

Number of Wolves Frequency

14  16 3

17  19 4

20  24 3

25  29 5

d. Wolves Observed

Number of Wolves Frequency

14  17 5

18  21 3

22  25 3

26  29 4

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Only the Fahrenheit one plz

Hoochie [10]

The formula given is:
F = 1.8C + 32
The temperature C is given as -20
So, all you have to do is plug the given C temperature in the formula to get the corresponding F temperature as follows:
F = 1.8(-20) + 32 = -36 + 32 = -4 degrees Fahrenheit

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3 years ago

What is the square root of 74 rounded to the nearest tenth

WARRIOR [948]

Answer:

8.6

Step-by-step explanation:

The square root of 74 is 8.602325, which rounded to the nearest tenth, is 8.6

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3 years ago

Write down the characteristics of culture translate into nepali language

Tatiana [17]

Answer:

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3 years ago

Answer questions a) and b)

Mariulka [41]

Answer:

$x\hookrightarrow -1\hookrightarrow 0\hookrightarrow 1\hookrightarrow 2\hookrightarrow 3\hookrightarrow 4\hookrightarrow 5$

$y\hookrightarrow 5\hookrightarrow 0\hookrightarrow -3\hookrightarrow -4\hookrightarrow -3\hookrightarrow 0\hookrightarrow 5$

The curve (B) matches the graph of y=x²-4x.

$----------\\Hope \ it\;helps$

4 0

3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

kupik [55]

Answer:

(a) 0.873 (b) 0.007 (c) 0.715 (d) 0.277 (e) 1.25 and 1.09

Step-by-step explanation:

The probability that the random variable X takes the value x is given by P(X=x) = $25Cx 0.05^x0.95^{25-x}$ . Then,

(a) $P(X\leq) = (25C0)(0.05^0)(0.95^{25}) + (25C1)(0.05^1)(0.95^{24}) + (25C2)(0.05^2)(0.95^{23})= 0.873$

(b) $P(X\geq5) = 1-P(X\leq4) = 1 - (0.873 + (25C3)(0.05^3)(0.95^{22}) + (25C4)(0.05^4)(0.95^{21})) = 1 - (0.873 + 0.12) = 0.007$

(c) $P(1\leqX\leq4) = (25C1)(0.05^1)(0.95^{24}) + (25C2)(0.05^2)(0.95^{23}) + (25C3)(0.05^3)(0.95^{22}) + (25C4)(0.05^4)(0.95^{21}) = 0.715$

(d) $P(X=0) = (25C0)(0.05^0)(0.95^{25}) = 0.277$

(e) E(X) = np = (25)(0.05) = 1.25 and $Sd(X) = \sqrt{Var(X)} = \sqrt{np(1-p)} = 1.09$

8 0

3 years ago