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3 years ago

Can you solve this riddle?

Mathematics

2 answers:

deff fn [24]3 years ago

4 0

Pair of shoes = 10
Man = 5
2 flowers =4
$5 + (5 + 4 + 10) \times 2 =$
$5 + (19) \times 2 =$
$5 + 38 =$
$43$

just olya [345]3 years ago

3 0

Shoes equal 10.

Man equals 5.

Flowers equal 3.

10 plus 5 times 3.

Pemdas.

10 plus 15.

the answer is 30.

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SNOW The first four snowfalls of the year in Shawnee's hometown measured 1.6 inches, 2.2 inches, 1.8 inches, and 1.4 inches. Use

wel

Answer:

The total amount of snow that fell is 7 inches.

Step-by-step explanation:

Given:

The first four snowfalls of the year in Shawnee's hometown measured 1.6 inches, 2.2 inches, 1.8 inches, and 1.4 inches.

Now, to find the total amount of snow that fell.

Shawnee's hometown measured first four snowfalls of the year are:

1.6 inches, 2.2 inches, 1.8 inches, and 1.4 inches.

Now, to get the total amount of snowfall we add all the four snowfalls of the year:

$1.6\ inches+2.2\ inches+1.8\ inches+1.4\ inches\\\\=7\ inches.$

Therefore, the total amount of snow that fell is 7 inches.

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4 years ago

An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total num

OverLord2011 [107]

Answer:

At the 3rd day 160 students caught flu.

Step-by-step explanation:

Consider the provided model P = − t ² + 13t + 130 where 1 ≤ t ≤ 6

The total number of students is represents by P.

We need to find the day that 160 students had the flu.

Substitute the value of P=160 in above formula.

$160 = -t^2 + 13t + 130$

$160+t^2 - 13t - 130=0$

$t^2 - 13t +30=0$

$t^2 - 10t-3t +30=0$

$t(t-10)-3(t-10)=0$

$(t-10)(t-3)=0$

Hence the value of t=10 or t=3

But it is given that 1 ≤ t ≤ 6, Therefore select t=3

Hence, at the 3rd day 160 students caught flu.

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3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7

Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable X ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.7 and standard deviation, σ = 0.75.

(a)

A random sample of n = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

$P(\bar X\leq 3.00)=P(\bar X$

$=P(\bar X$

$=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$=P(Z$

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

$P(2.70$

$=P(0$

*Use a z-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

$P(\bar X\leq 3.00)\geq 0.99$

$P(\bar X$

$P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$P(Z$

The value of z for the probability above is

z ≥ 2.33

Compute the value of n as follows:

$\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33$

$\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}$

$\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77$

Thus, the sample size must be at least 77.

8 0

3 years ago

Simplify: 3(a-2)+4(3-a)-10

Neporo4naja [7]

Answer:

-a - 4

Step-by-step explanation:

3(a-2)+4(3-a)-10

3a - 6 + 12 - 4a - 10

-a - 4

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4 years ago

Explain how to find the upper quartile of a box-and-whisker plot.

kirza4 [7]

B. The upper quartile is the median of the upper half of the data.

if I am wrong please let me know.

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3 years ago

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