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3 years ago

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Jenny's company has 15 trucks in two different sizes. There are small trucks with 6 wheels and large trucks with 12 wheels. Her

company needs to inspect the 108 wheels of all of the trucks. How many large trucks does the company have?

2 answers:

stira [4]3 years ago

6 0

Answer: 8

Step-by-step explanation:

Vladimir79 [104]3 years ago

5 0

Answer: 9

Step-by-step explanation:

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A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri

rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X $\sim$ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z $\sim$ N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

<em>μ</em> = 60 minutes

<em>σ </em> = 12 minutes

Compute the <em>z</em>-score for the student being active 48 minutes as follows:

$Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0$

Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.

The correct option is (b).

4 0

3 years ago

postnew [5]

Answer:

Below

Step-by-step explanation:

Substituting the given values:

f(6) = 6(2/3) - 2 = cube root of 6^2 - 2 = cube root 36 - 2

f(-6)= (-6)(2/3) - 2 = cube root of(-6)^2 - 2 = cube root 36 - 2

So This is true,

f(6) = cube root of 6^2 - 2 = cube root 36 - 2 = 1.3019

2 * f(3) = 2 * (cube root of 3^2 - 2 ) = 2 * (cube root of 9 - 2) = 0.1602

So False,

4 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago

If John makes 30dollars/hour how much will he make in a year

stich3 [128]

I... don’t know? How many hours is he working a year? How many days is he working?

5 0

3 years ago

Read 2 more answers

Does it matter which two points you use to create a slope triangle?

Fynjy0 [20]

Answer:

yes

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers