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3 years ago

PLEASEEEE Write an expression that contains at least two terms and is in simplest form.

Mathematics

1 answer:

never [62]3 years ago

7 0

The answer is 6m - 6 = 8m + 4

Hope this helps please give brainless!!!

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10/12 + 1/4 please help me

BartSMP [9]

Answer:

Here you go, hope this helps

Step-by-step explanation:

Exact form: 13/12

Decimal form: 1.083 repeating

Mixed Number form: 1 1/12

3 0

3 years ago

Read 2 more answers

W−6 =−4w−11 <br> what is w?

ziro4ka [17]

Answer:

w=-1

Step-by-step explanation:

w-6=-4w-11

5w=-5

w=-1

7 0

3 years ago

Read 2 more answers

A woman is randomly selected from the 18–24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

Naddik [55]

Answer:

$X \sim N(114.8,13.1)$

Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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:

And we can find this probability using the complement rule:

$P(z>1.924)=1-P(z$

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".

Solution to the problem

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

$X \sim N(114.8,13.1)$

Where $\mu=114.8$ and $\sigma=13.1$

We are interested on this probability

$P(X>140)$

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: $P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140- 1114.8}{2.6})=P(z>1.924)$ And we can find this probability using the complement rule: