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3 years ago

Show how to solve 1/4 (4 + x) = 4/3 pls solve for 2 brainliest

1 answer:

4 0

Simplify
1
4
(
4
+
x
)
4
1

(4+x) to
4
+
x
4
4
4+x

.
4
+
x
4
=
4
3
4
4+x

=
3
4

2 Simplify
4
+
x
4
4
4+x

to
1
+
x
4
1+
4
x

.
1
+
x
4
=
4
3
1+
4
x

=
3
4

3 Subtract
1
1 from both sides.
x
4
=
4
3
−
1
4
x

=
3
4

−1

4 Simplify
4
3
−
1
3
4

−1 to
1
3
3
1

.
x
4
=
1
3
4
x

=
3
1

5 Multiply both sides by
4
4.
x
=
1
3
×
4
x=
3
1

×4

6 Simplify
1
3
×
4
3
1

×4 to
4
3
3
4

.
x
=
4
3
x=
3
4

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3 years ago

EXPAND:

Sladkaya [172]

Answer:

$\displaystyle A) 5\log( {x}^{} ) + 2 \log( {y}^{} )$

Step-by-step explanation:

we would like to expand the following logarithmic expression:

$\displaystyle \log( {x}^{5} {y}^{2} )$

remember the multiplication logarithmic indentity given by:

$\displaystyle \rm \log( \alpha \times \beta ) \iff \log( \alpha ) + \log( \beta )$

so our given expression should be

$\displaystyle \log( {x}^{5} ) + \log( {y}^{2} )$

by exponent logarithmic property we acquire:

$\displaystyle 5\log( {x}^{} ) + 2 \log( {y}^{} )$

hence, our answer is A

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3 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

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

$P( z>2.143) = 1-P(Z$

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

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

$P( z>2.143) = 1-P(Z$

4 0

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dybincka [34]

The second option! 16, 20, 24, 28 can all be divided by 4

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3 years ago

A pet shop manager wants the same number of birds in each cage he wants to use as few cages as possible but he can only have one

anastassius [24]

Since there are more parakeets than canaries, it is not possible to have only 1 of each bird in each cage <u>and</u> have the same number of birds in each cage.

He could use 42 cages, putting a canary in with the parakeet in 18 of them. Then he would have 18 cages with 2 birds each, and 24 cages with 1 bird each.

The only way to have the same number of birds (1) in all cages is to have 60 cages, 42 of which have 1 parakeet, and 18 of which have 1 canary.

_____

If more than 1 of each kind of bird can be put in the cage, the collection of birds could be put into 6 cages, each of which would be home to 7 parakeets and 3 canaries.

4 0

3 years ago