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3 years ago

How do you solve this problem in the photo?

Mathematics

2 answers:

Klio2033 [76]3 years ago

6 0

Answer: 5^2

Step-by-step explanation:

Since the base are all the same, you will add the exponents and let the exponents stay the same.

5^3 * 5^4= 5^7 * 5^-5= 5^2

Mashcka [7]3 years ago

6 0

Answer:

Step-by-step explanation:

Using the rule of exponents

$a^{m}$ × $a^{n}$ ⇔ $a^{(m+n)}$

The rule can be extended to include more than 3 terms.

Given

5³ × $5^{4}$ × $5^{-5}$

= $5^{(3+4-5)}$

= 5²

= 25

You might be interested in

The legs of an isosceles right triangle both measure 10 inches. Find the length of the hypotenuse.

gavmur [86]

Answer:

14.14

Step-by-step explanation:

10^2 + 10^2 = c^2

100 + 100 = c^2

200 = c^2

c = 14.14

3 0

3 years ago

SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as

vovangra [49]

Answer:

a) $P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

$P(Z>1.25)=1-P(Z$

b) $P(400$

$P(-1$

c) $z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

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 SAT scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(500,100)$

Where $\mu=500$ and $\sigma=100$

We are interested on this probability

$P(X>625)$

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:

$P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

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

$P(Z>1.25)=1-P(Z$

Part b

We are interested on this probability

$P(400$

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:

$P(400$

And we can find this probability with this difference:

$P(-1$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-1$

Part c

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

$P(X>a)=0.8$ (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

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

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

8 0

3 years ago

Ayden's monthly bank statement showed the following deposits and withdrawals:

kumpel [21]

$65.91 should be the account balance at the end of the month

3 0

3 years ago

If a TV has 200 ft by 875 ft what is the total mass of it

mafiozo [28]

There's no way to find the mass with the information given.
"200 ft" and "875 ft" are linear dimensions ... things you would
measure with a ruler. They tell us how long or wide or high the
TV is. None of that information tells us anything about its mass.

By the way ... those are gigantic measurements. I was working
in Wrigley Field here in Chicago just before opening day in 2015,
when the new video screens were being installed. I saw them
up close while the cranes lifted them into position. None of the
Chicago Cubs' new screens is as big as the numbers in the question,
and I'm pretty sure the whole score-board isn't that big either.
Something is definitely wrong here.

8 0

3 years ago

Read 2 more answers

Here is a question for you

dimaraw [331]

Answer:

3/8

Step-by-step explanation:

they simply combined the like terms (aka the x numbers in this case) so all that's left is 3/8

6 0

3 years ago