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2 years ago

5

A motorist has gotten a flat tire and is in the process of changing it. One of the lug nuts is on very tight and he is trying to

remove it

1 answer:

Marat540 [252]2 years ago

5 0

The best approach remove the nut is to fine a tool or material that that can help loosen the nut fast.

<h3>What are the steps to change a flat tire?</h3>

People sometimes are often stuck with a flat tire and this can be disheartening. The best way to go about it is to:

Find a good Place and then pull over.
Always use the Hazard Lights and Parking Brake.
Check for Materials that you need to change the tires.
Loosen the Lug Nuts and then begin the changing.

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Which takes place during the courtship stage of the family life cycle? A. A child becomes an adult. B. People meet and fall in l

tigry1 [53]

The answer is B. <span>. People meet and fall in love. </span>
Courtship stage of the family life cycle happens before the couple entered the marriage stage.
During this stage, couples will made various effort to attract each other (flirting or giving gifts) and may convince themselves that they're finally found the 'right one'

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3 years ago

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If f(x)=(x+3)³+4<br> let g(x)=f(x+1)-2<br> find when g(x)=12

Alla [95]

Answer:

Explanation:

g(x) = f(x + 1) - 2

g(x) = (x + 1 + 3)^3 -2

g(x) = (x + 4)^3 - 2

Now what are we supposed to do with this? I'm guessing we are to find what value of x will make g(x) = 12. If I'm wrong, leave a note in the form of a comment.

g(x) = 12

(x + 4)^3 - 2 = 12 Add 2 to both sides.

(x + 4)^3 = 14 Take the cube root of both sides.

x + 4 = cubrt(14)

x + 4 = 2.4101 Subtract 4

x = 2.4101 - 4

x = -1.5899

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2 years ago

2. Which of the following infractions is worth 3

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Having a child restraint violation is 3 points off your license

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3 years ago

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a researcher is studying a group of field mice. the distribution of the weight of field mice is approximately normal with mean 2

yarga [219]

The proportion of field mice weighing more than 33 grams in the distribution will be about 0.2275

Mean, μ = 25 grams
Standard deviation, σ = 4 grams
Score, X = 33 grams

We need to calculate the P(Z > Zscore)

Recall :

Zscore = (X - μ) ÷ σ

For X > 33 :

Zscore = (33 - 25) ÷ 4 = 2

Therefore, the probability would be expressed as :

P(Z > 2) = 1 - P(Z < 2)

From the normal distribution table :

P(Z < 2) = 0.97725

Therefore,

P(Z > 2) = 1 - 0.97725 = 0.02275

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1 year ago

company manufactured six television sets on a given day, and these TV sets were inspected for being good or defective. The resul

lana [24]

Sampling distribution involves the proportions of a data element in a given sample.

<em>The proportion of Good TV set is 0.67</em>
<em>The number of ways of selecting 5 from 6 TV sets is 6</em>
<em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

$p = \frac{4}{6}$

$p = 0.67$

<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

$^6C_5 = \frac{6}{1}$

$^6C_5 = 6$

Hence, there are 6 ways

<u>(c) All possible sample space of 4</u>

First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

$^6C_4 = 15$

So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

The proportion column is calculated by dividing the number of Good TVs by the total selected (4) i.e.

$p = \frac{Good}{n}$

<u>(d) The sampling distribution</u>

In (a), we have:

$p = 0.67$ --- proportion of Good TV

The sampling error is calculated as follows:

$SE_n = |p - p_n|$

So, we have:

$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

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3 years ago