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3 years ago

Is the expression 70 – (4 • 15) a difference of three terms, a product of two terms, or a difference of two terms?

Mathematics

1 answer:

nlexa [21]3 years ago

6 0

Answer:

7 4 and 15 because term means numbers

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galina1969 [7]

all you will have to do is add the top number . you should D

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4 years ago

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If a figure is translated 5 units down and 3 units right, how do you represent it algebraically

qwelly [4]

Answer

(x+3) -5

Step-by-step explanation:

plug the values into the parent transformation equation to its rightful places

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3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

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3 years ago

On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the d

Reika [66]

Answer:

P ( -1 < Z < 1 ) = 68%

Step-by-step explanation:

Given:-

- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :

u = 67.2

s.d = 4.6

- The random variable (X) that denotes standardized test scores following normal distribution:

X~ N ( 67.2 , 4.6^2 )

Find:-

What percent of the data fell between 62.6 and 71.8?

Solution:-

- We will first compute the Z-value for the given points 62.6 and 71.8:

P ( 62.6 < X < 71.8 )

P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )

P ( -1 < Z < 1 )

- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:

P ( -1 < Z < 1 ) = 68%

P ( -2 < Z < 2 ) = 95%

P ( -3 < Z < 3 ) = 99.7%

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3 years ago

What is the volume of the cone

deff fn [24]

292 is teh answer ______

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3 years ago

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