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Nitella [24]
2 years ago
12

What is 5 dogs + 180 cats?

Mathematics
2 answers:
shtirl [24]2 years ago
7 0

Answer:

Step-by-step explanation:

185 dogs and cats

aleksley [76]2 years ago
3 0

Answer:

185 animals altogether ig?

Step-by-step explanation:

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I literally don’t get this I need the answer
Ksju [112]

Answer:

62.83 units^{3}

Step-by-step explanation:

The equation is V(Volume) equals π (Pi/3.14) multiplied by r(radius) squared multiplied by h(height). The way it would be written out is V=π×r^{2}×h. So when you plug in the radius (2) and the height (5) you get π×2^{2}×5 which equals 62.83185 or 62.83. I hope this helps you understand and if you need more help let me know.

6 0
2 years ago
If triangle ABC is similar to triangle PQR, find the value of x and y.
tester [92]

Answer:

x = 9

y = 6

Step-by-step explanation:

ΔABC ~ ΔPQR

Therefore, corresponding sides of these similar triangles will be in the proportional.

\frac{AB}{PQ}=\frac{BC}{QR}= \frac{AC}{PR}

\frac{6}{x}= \frac{4}{y}= \frac{8}{12}

\frac{6}{x}= \frac{8}{12}

x = \frac{12\times 6}{8}

x = 9

Similarly, \frac{4}{y}= \frac{8}{12}

y = \frac{12\times 4}{8}

y = 6

6 0
2 years ago
Can someone answer this its just boxplots :)))
777dan777 [17]
Nuclear engineer .......
8 0
2 years ago
If DAB = 80°, then DAC =
vodka [1.7K]

Answer:

80

Step-by-step explanation:

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6 0
1 year ago
Read 2 more answers
The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day
Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(between 236 and 281 days)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%

b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

1-\dfrac{1}{(2)^2} = 75\%

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0
2 years ago
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