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Samuel solved the equation h-234=1,192 and found the solution to be h=958. Is he correct? If not, why not?

Naya [18.7K]

Answer:

no he is not correct because he did not add 234 and 1192 to find the value of h, which is 1426

Step-by-step explanation:

h-234 = 1192

add 234 to each side

h-234+234= 1192+234

h = 1192+234

h = 1426

4 0

3 years ago

Read 2 more answers

4.9t+0.3=5.6t+0.72 I need help on my work lol

Whitepunk [10]

The first step that we must take before attempting to solve the problem is to understand what the problem is asking us to do and what is given to us to help accomplish that goal. Although it does not explicitly state that we must solve for t, this is usually what the problem statement would be asking if we just receive and expression like this. What is given to us to accomplish that goal is the expression $4.9t+0.3=5.6t+0.72$ .

Now that we have completed that step, we can move onto the next part which is actually solving the problem. The next step that we should take when solving for the unknown, in this case t, is to subtract 4.9t from both sides.

Subtract 4.9t from both sides

$4.9t+0.3=5.6t+0.72$

$(4.9t-4.9t)+0.3=(5.6t-4.9t)+0.72$

$0.3=(5.6t-4.9t)+0.72$

$0.3=(0.7t)+0.72$

Now that we got all of the t's to one side, let us isolate t completely and the next step that we should take is to subtract 0.72 from both sides.

Subtract 0.72 from both sides

$0.3=0.7t+0.72$

$0.3 - 0.72=0.7t+0.72 - 0.72$

$0.3 - 0.72=0.7t$

$-0.42=0.7t$

The final step that we need to take to isolate t would be to divide both sides by 0.7 which would remove the coefficient from the unknown variable t and divide 0.7 from -0.42

Divide both sides by 0.7

$-0.42=0.7t$

$\frac{-0.42}{0.7}=\frac{0.7t}{0.7}$

$\frac{-0.42}{0.7}=t$

$-0.6=t$

Therefore, after fully narrowing down the solution we were able to determine that the solution of the unknown variable or t is equal to -0.6

5 0

2 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$