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liraira [26]
3 years ago
9

Question is shown below.

Mathematics
1 answer:
Andrei [34K]3 years ago
3 0

ANSWER

D. 4 minutes 40 seconds

EXPLANATION

The given data set showing the number of seconds of each song the playlist is:

283,204,198,229,173,230,168,236,215,204

Let the length of the 11th song be x.

Because he wants the mean to be 3 minutes 40 seconds, which is equal to 220 seconds, we write and solve the following equation for x.

\frac{283 + 204 + 198 + 229 + 173 + 230 + 168 + 236 + 215 + 204 + x}{11}  = 220

\frac{2140 + x}{11}  = 220

2140 + x = 2420

x = 2420 - 2140

x = 280

We convert 280 seconds to minutes.

\frac{280}{60}  = 4 \frac{2}{3}  = 4mins : 40 sec

The correct choice is D

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Find the length of the diagonal of a rectangular football pitch with sides 46.6 m and 78.9 m.
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Answer:

d≈91.63m

Step-by-step explanation:

d=w2+l2=78.92+46.62≈91.63389m

7 0
3 years ago
Given the information in the diagram below, find the measure of angle N. Use RACE and complete sentences to show how you found t
Sonja [21]

Answer:

m<N = 76°

Step-by-step explanation:

Given:

∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).

m<J = 64° (given)

Required:

m<N

SOLUTION:

m<K = m<J (base angles of an isosceles ∆ are equal)

m<K = 64° (Substitution)

m<K + m<J + m<JLK = 180° (sum of ∆)

64° + 64° + m<JLK = 180° (substitution)

128° + m<JLK = 180°

subtract 128 from each side

m<JLK = 180° - 128°

m<JLK = 52°

In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.

Thus:

m<MLN = m<JKL (vertical angles are congruent)

m<MLN = 52°

m<M = m<MLN (base angles of isosceles ∆MNL)

m<M = 52° (substitution)

m<N + m<M° + m<MLN = 180° (Sum of ∆)

m<N + 52° + 52° = 180° (Substitution)

m<N + 104° = 180°

subtract 104 from each side

m<N = 180° - 104°

m<N = 76°

4 0
3 years ago
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Answer:

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Step-by-step explanation:

8 0
3 years ago
A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The
TiliK225 [7]

Correct question:

A food company sells salmon to various customers. The mean weight of the salmon is 44 lb with a standard deviation of 3 lbs. The company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 16 ​salmon, and to discount outlet stores in pallets of 64 salmon. To forecast​ costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts​ (a) and​ (b) below.

a. Find the standard deviations of the mean weight of the salmon in each type of shipment.

b. The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes or pallets?

Answer:

Given:

Mean, u = 44

Sd = 3

The company ships in boxes of 9, cartons of 16 and pallets of 64.

a) For the standard deviations of the mean weight of the salmon in each type of shipment, lets use the formula: \frac{s.d}{\sqrt{u}}

i) For the standard deviation of the mean weight of salmon in boxes of 9, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{9}}

= \frac{3}{3} = 1

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ii) For the standard deviation of the mean weight of salmon in cartons of 16, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{16}}

= \frac{3}{4} = 0.75

Standard deviation = 0.75

iii) For the standard deviation of the mean weight of salmon in pellets of 64, we have:

\frac{s.d}{\sqrt{u}}

= \frac{3}{\sqrt{64}}

= \frac{3}{8} = 0.375

Standard deviation = 0.375

b) The distribution of shipping weights would be better characterized by a Normal model for the pallets, because regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample increases.

5 0
3 years ago
Solve for x and y please help!!
Pie

Answer:

the answer is

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5 0
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