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3 years ago

If your brakes suddenly give out and your car is about to hit a tree, what should you never do?

2 answers:

8 0

Depending on where the tree is car is located at, it is best to try not to move as much, unless you are extremely sure you are able to get out without falling off a cliff or something. If you are not sure that you can make it out, try to contact 9-1-1 and tell them your location and whats the trouble. After your call, just wait until reinforcements arrive, excuse me i meant rescue lol

hope this helps XD

inysia [295]3 years ago

4 0

The one thing you should never do is go faster

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What is -0.4 as a decimal?

evablogger [386]

Decimal. Decimal is the base-10 notational system for representing real numbers. The expression of a number using the decimal system is called its decimal expansion, examples of which include 1, 13, 2028, 12.1, and 3.14159.

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3 years ago

Read 2 more answers

2. The actual tree on the left is 10 feet tall. The model tree on the right is 8 inches tall. What is the scale factor?

Lina20 [59]

Answer:

D. 1/15

Step-by-step explanation:

Actual tree = 10 ft = 12 × 10 = 120 in.

Model tree = 8 in.

Scale factor = model tree/actual tree

Scale factor = 8/120

Scale factor = 1/15

4 0

2 years ago

Mark owns Siberian Husky sled dogs. He knows from data collected over the years that the weight of the dogs is a normal distribu

kozerog [31]

Calculate the z-score for the given data points in the item using the equation,

z-score = (x - μ) / σ

where x is the data point, μ is the mean, and σ is the standard deviation.

Substituting,
(47.7) z-score = (47.7 - 52.5)/2.4 = -2

This translates to a percentile of 2.28%.

(54.9) z-score = (54.9 - 52.5)/2.4 = 1

This translates to a percentile of 84.13%.

Then, subtract the calculate percentiles to give us the final answer of 81.85%.

Thus, 81.85% of the Siberian Husky sled dogs are expected to weigh between 47.7 and 54.9 lbs.

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3 years ago

Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random

Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = $50,000

$\sigma$ = population standard deviation = $25,000

n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P( $\bar X$ > $52,000)

P( $\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

= 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

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3 years ago

How to write five and nine tenths in standard from

Trava [24]

5 9/10 .........................................

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3 years ago