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3 years ago

11

In two or more complete sentences, use your knowledge of the unit circle to describe why the following trigonometric

1 answer:

JulsSmile [24]3 years ago

7 0

Should be 59.490 just because if you have to round the tenth and the hundredth together

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The value of the cumulative standardized normal distribution at z is 0.8770. what is the value of z?

Vinvika [58]

<span>The value of z is 0.8770. This answer was derived from information in the question, which states that the value of z is 0.8770. Granted, I don't understand what is meant by "cumulative standardized normal distribution." I only passed statistics because the professor felt sorry for me.</span>

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3 years ago

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Line l was mapped to Line m as shown in the graph below. please help

steposvetlana [31]

Line I was translate up 5 units

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3 years ago

please help thank you

Free_Kalibri [48]

k(-3)=36(2)

hope it helps

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2 years ago

can u tell me which letter the answer is.. i know how to do this problem im just confused what the answer is

Ede4ka [16]

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Im pretty sure its a :3

Step-by-step explanation:

7 0

2 years ago

On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes

belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

$r = \frac{\Delta T_{X}}{\Delta T_{Y}}$

$r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}$

$r = 11\,\frac{^{\circ}X}{^{\circ}Y}$

The difference between current temperature in Y linear scale with respect to freezing point is:

$\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)$

$\Delta T_{Y} = 115\,^{\circ}Y$

The change in X linear scale is:

$\Delta T_{X} = r\cdot \Delta T_{Y}$

$\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)$

$\Delta T_{X} = 1265\,^{\circ}X$

Lastly, the current temperature on the X scale is:

$T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X$

$T_{X} = 1150\,^{\circ}X$

The current temperature on the X scale is 1150 °X.

5 0

3 years ago