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3 years ago

What it the unit rate of 183 miles to 3 hours

Mathematics

2 answers:

Sophie [7]3 years ago

7 0

Answer:

61 mph

Step-by-step explanation:

so if you go 183 miles in 3 hours

183 ÷ 3= 61

61 miles every hour

Over [174]3 years ago

4 0

Answer:

61 miles per hour

Step-by-step explanation:

you divide 183 by 3 to get 61, so the unit rate is 61 miles per hour

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You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass (3,000 kg each)

Kazeer [188]

Answer:

1) Kinetic energy of the truck 1 is 600,000 joules.

Kinetic energy of the truck 2 is 1,837,500 joules.

2 a)Potential energy of truck 1 is 646,800 Joules.

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

Step-by-step explanation:

Mass of the truck 1 and truck 2,m = 3000 kg

Velocity of the truck 1, $v_1$ = = 20 m./s

Velocity of the truck 2, $v_2$ = 35 m./s

1. Kinetic energy of the truck 1: $\frac{1}{2}mv_{1}^2=\frac{1}{2}\times 3000 kg\times (20 m/s)^=600,000 joules$

Kinetic energy of the truck 2: $\frac{1}{2}mv_{2}^2=\frac{1}{2}\times 3000 kg\times (35 m/s)^=1,837,500 joules$

2.a) Energy loose by the truck 1:

Initial height of the truck on the hill = 22 m

At that height potential energy will be maximum:

Potential energy of truck 1 at that height = $mgh=3000\times 9.8 m/s^2\times 22 m=646,800 Joules$

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

3 0

3 years ago

Math problem algebra

emmasim [6.3K]

My friend the answer is calculated by dividing denominator by 2 ie the answer is 1.

5 0

3 years ago

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of

jarptica [38.1K]

Answer:

The "probability that a given score is less than negative 0.84" is $\\ P(z$ .

Step-by-step explanation:

From the question, we have:

The random variable is normally distributed according to a standard normal distribution, that is, a normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .
We are provided with a z-score of -0.84 or $\\ z = -0.84$ .

Preliminaries

A z-score is a standardized value, i.e., one that we can obtain using the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

x is the raw value coming from a normal distribution that we want to standardize.
And we already know that $\\ \mu$ and $\\ \sigma$ are the mean and the standard deviation, respectively, of the normal distribution.

A z-score represents the distance from $\\ \mu$ in standard deviations units. When the value for z is negative, it "tells us" that the raw score is below $\\ \mu$ . Conversely, when the z-score is positive, the standardized raw score, x, is above the mean, $\\ \mu$ .

Solving the question

We already know that $\\ z = -0.84$ or that the standardized value for a raw score, x, is below $\\ \mu$ in 0.84 standard deviations.

The values for probabilities of the standard normal distribution are tabulated in the standard normal table, which is available in Statistics books or on the Internet and is generally in cumulative probabilities from negative infinity, - $\\ \infty$ , to the z-score of interest.

Well, to solve the question, we need to consult the standard normal table for $\\ z = -0.84$ . For this:

Find the cumulative standard normal table.
In the first column of the table, use -0.8 as an entry.
Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
The intersection of these two numbers "gives us" the cumulative probability for z or $\\ P(z$ .

Therefore, we obtain $\\ P(z$ for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the standard normal distribution, $\\ N(0,1)$ , at the left of z = -0.84.

To "draw a sketch of the region", we need to draw a normal distribution (symmetrical bell-shaped distribution), with mean that equals 0 at the middle of the distribution, $\\ \mu = 0$ , and a standard deviation that equals 1, $\\ \sigma = 1$ .

Then, divide the abscissas axis (horizontal axis) into equal parts of one standard deviation from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, $\\ -\sigma$ , from it). All the area to the left of this value must be shaded because it represents $\\ P(z$ and that is it.