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2 years ago

In the picture, the angle made by the goniometer is classified as a(n) _ angle. _ ° < θ < ___ .

Mathematics

3 answers:

Svet_ta [14]2 years ago

5 0

An obtuse angle which is greater than 90 degrees and less than 180 degrees.

bixtya [17]2 years ago

4 0

Answer:

In the picture, the angle made by the goniometer is classified as a(n) obtuse angle.

90° < θ < 180°

1. obtuse

2. 90°

3. 180°

Guest1 year ago

0 0

Idk the answer

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The graph shows the distance in miles of a runner over x hours. What is the average rate of speed over the interval [9, 11]?

JulsSmile [24]

For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
Therefore, we can find the average rate of change as follows:
m = (y2-y1) / (x2-x1)
m = (11-6) / (11-9)
m = (5) / (2)
m = 5/2
Answer:
the average rate of speed over the interval [9, 11] is:
D. 5 / 2

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2 years ago

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A movie theater is offering a special summer pass. Passholders pay $8 per movie for the first 5 movies and watch additional movi

Brrunno [24]

Number of movies —— C

1 —— c= 8
3——- c= 24
5—— c= 40
6 —— c = 40
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2 years ago

For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 52

Ludmilka [50]

Answer:

the probability that a sample of the 35 exams will have a mean score of 518 or more is 0.934 or 93.4%.

Step-by-step explanation:

This is s z-test because we have been given a sample that is large (greater than 30) and also a standard deviation. The z-test compares sample results and normal distributions. Therefore, the z-statistic is:

(520 - 518) / (180/√35)

= 0.0657

Therefore, the probability is:

P(X ≥ 0.0657) = 1 - P(X < 0.0657)

where

X is the value to be standardised

Thus,

P(X ≥ 0.0657) = 1 - (520 - 518) / (180/√35)

= 1 - 0.0657

= 0.934

Therefore, the probability that a sample of the 35 exams will have a mean score of 518 or more is 0.934 or 93.4%.

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2 years ago

Jimmy’s Delicatessen sells large tins of Tom Tucker’s Toffee. The deli uses a periodic review system, checking inventory levels

Yakvenalex [24]

Answer:

The restocking level is 113 tins.

Step-by-step explanation:

Let the random variable X represents the restocking level.

The average demand during the reorder period and order lead time (13 days) is, μ = 91 tins.

The standard deviation of demand during this same 13- day period is, σ = 17 tins.

The service level that is desired is, 90%.

Compute the z-value for 90% desired service level as follows:

$z_{\alpha}=z_{0.10}=1.282$

*Use a z-table for the value.

The expression representing the restocking level is:

$X=\mu +z \sigma$

Compute the restocking level for a 90% desired service level as follows:

$X=\mu +z \sigma$

$=91+(1.282\times 17)\\=91+21.794\\=112.794\\\approx 113$

Thus, the restocking level is 113 tins.

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2 years ago

The volume V, in liters, of air in the lungs during a four-second respiratory cycle is approximated by the model V = 0.1729t + 0

Roman55 [17]

It was stated that the respiratory cycle takes 4 seconds. So, you just have to simply substitute t=4 to the equation to find the average volume of air in the lungs per cycle.

$V =0.1729t+ 0.1522t^{2} -0.0374 t^{3}$

$V =0.1729(4)+ 0.1522(4^{2}) -0.0374 (4^{3} )$

V = 0.7332 liters of air in one cycle

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2 years ago

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In the picture, the angle made by the goniometer is classified as a(n) ___ angle. ___ ° < θ < ___ .

In the picture, the angle made by the goniometer is classified as a(n) _ angle. _ ° < θ < ___ .