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2 years ago

5-(6- 9x)=1-(4-10 ) please

Mathematics

1 answer:

MAVERICK [17]2 years ago

4 0

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Let's solve ~

$\qquad \sf \dashrightarrow \:5 - ( 6 - 9x) = 1 - (4 - 10)$

$\qquad \sf \dashrightarrow \:5 - 6 + 9x = 1 - 4 + 10$

$\qquad \sf \dashrightarrow \: 9x - 1 = 7$

$\qquad \sf \dashrightarrow \:9x = 8$

$\qquad \sf \dashrightarrow \:x = \dfrac{8}{9}$

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A gumball has a diameter that is 66 mm. The diameter of the gumball's spherical hollow core is 58 mm. What is the volume of the

grigory [225]

Answer:

Volume of gumball without including its hollow core is 48347.6 cubic mm.

Step-by-step explanation:

Given:

Diameter of Gumball = 66 mm

Since radius is half of diameter.

Radius of gumball = $\frac{diameter}{2}=\frac{66}{2} =33 \ mm$

Now We will first find the Volume of Gumball.

To find the Volume of Gumball we will use volume of sphere which is given as;

Volume of Sphere = $\frac{4}{3}\pi r^3$

Now Volume of Gumball = $\frac{4}{3}\times3.14 \times (33)^3 = 150456.24 \ mm^3$

Also Given

Diameter of gumball's spherical hollow core = 58 mm

Since radius is half of diameter.

Radius of gumball's spherical hollow core = $\frac{diameter}{2}=\frac{58}{2} =29 \ mm$

Now We will find the Volume of gumball's spherical hollow core.

Volume of Sphere = $\frac{4}{3}\pi r^3$

So Volume of gumball's spherical hollow core = $\frac{4}{3}\times3.14 \times (29)^3 = 102108.61 \ mm^3$

Now We need to find volume of the gumball without including its hollow core.

So, To find volume of the gumball without including its hollow core we would Subtract Volume of gumball spherical hollow core from Volume of Gumball.

volume of the gumball without including its hollow core = Volume of Gumball - Volume of gumball's spherical hollow core = $150456.24\ mm^3 - 102108.61\ mm^3 = 48347.63\ mm^3$

Rounding to nearest tenth we get;

volume of the gumball without including its hollow core = $48347.6\ mm^3$

Hence Volume of gumball without including its hollow core is 48347.6 cubic mm.

8 0

3 years ago

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob

Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that $\mu = 36, \sigma = 5.5$

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that $n = 9$ . So

$s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333$

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{38 - 36}{1.8333}$

$Z = 1.09$

$Z = 1.09$ has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{34.5 - 36}{1.8333}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

4 0

3 years ago

Prove that sinxtanx=1/cosx - cosx

maks197457 [2]

Answer:

See below

Step-by-step explanation:

We want to prove that

$\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}$

Taking the RHS, note

$\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}$

Remember that

$\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)$

Therefore,

$\dfrac{1-\cos^2(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)} = \dfrac{\sin(x)\sin(x)}{\cos(x)}$

Once

$\dfrac{\sin(x)}{\cos(x)} = \tan(x)$

Then,

$\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)$

Hence, it is proved

5 0

3 years ago

Thank you in advance for the help !!

babymother [125]

The Answer is C, my good friend.

4 0

3 years ago

List the following numbers from least to greatest -5/3, 7/3, -3/4, 7/5.

sukhopar [10]

-3/4, -5/3, 7/5, 7/3

6 0

3 years ago

5-(6- 9x)=1-(4-10 ) please​

5-(6- 9x)=1-(4-10 ) please