Help with this algebra 2 question

How do I apply properties of multiplication powers ? Answer in 4 sentences or more.

Zielflug [23.3K]

To<span> find a </span>power of a product, look the power of each factor and then multiply. If you want to multiply two powers with the same exponent but different bases, you use the commutative property of multiplication. Product of powers property shows us that when you multiply powers with the same base you just have to add the exponents. This is raising a <span>power to a power.</span>

4 0

3 years ago

The most common women’s shoe size in the U.S. is reported to be an 8 1/2 . A shoe store uses a table like the one below to decid

vesna_86 [32]

Answer:

the answer is caca

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Evaluate x(y-z)2 for =-1,y=5,and z=

Nata [24]

Answer:

The value of $x(y-z)^2$ is -16.

Step-by-step explanation:

We need to evaluate $x(y-z)^2$ for x = -1, y = 5 and let us assume that z = 1

It can be simply done by putting the values of x,y and z in the given expression.

$x(y-z)^2=(-1)(5-1)^2\\\\=-1(4)^2$

We know that the value of 4² = 16

So,

$x(y-z)^2=16\times -1\\\\=-16$

So, the value of $x(y-z)^2$ is -16.

6 0

3 years ago

Avoiding an accident while driving can depend on reaction time. Suppose that reaction time, measured from the time the driver fi

Nadya [2.5K]

Answer:

2.5% of drivers have a reaction time more than 2.14 seconds

16% of drivers have a reaction time less than 1.78 seconds

84% of drivers have a reaction time less than 2.02 seconds

Step-by-step explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

The normal distribution is symmetric, which means that 50% of the measures are below the mean and 50% are above.

In this problem, we have that:

Mean = 1.9s

Standard deviation = 0.12

What percentage of drivers have a reaction time more than 2.14 seconds?

2.14 = 1.9 + 2*0.12

So 2.14 is two standard deviations above the mean.

Of the 50% of the measures above the mean, 95% are within 2 standard deviations of the mean, so, below 2.14. The other 5% is above.

0.05*0.5 = 0.025

2.5% of drivers have a reaction time more than 2.14 seconds

What percentage of drivers have a reaction time less than 1.78 seconds?

1.78 = 1.9 - 0.12

So 1.78 is one standard deviation below the mean.

Of the 50% of the measures that are below the mean, 68% are within one standard deviation of the mean, that is, greater than 1.78.

100 - 68 = 32

0.32*50 = 0.16

16% of drivers have a reaction time less than 1.78 seconds

What percentage of drivers have a reaction time less than 2.02 seconds?

2.02 = 1.9 + 0.12

So 2.02 is one standard deviation above the mean.

Of the measures that are below the mean, all are below 2.02.

Of those that are above, 68% are below 2.02.

0.5 + 0.68*0.5 = 0.84

84% of drivers have a reaction time less than 2.02 seconds

5 0

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

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

3 0

3 years ago