ALGEBRA QUESTION<br> x+7=8.3;x=1.3 TRUE OR FALSE

Don't be childish I will report thanks

Semenov [28]

Answer:

what do you mean childish?.. anyway your answer is

-2| 3 - 6 |

= -6

happy to help!

5 0

3 years ago

Read 2 more answers

Will give brainliest

ehidna [41]

Answer:

25.047 or roughly 25.

Step-by-step explanation:

I solved it wrong the first time then i double check with wolframalpha and got the correct number.

We know that the area of the tile is 18 $\sqrt{3}$ and it makes a triangle, we also know the base and height. In this case the base is 2 $\sqrt{3}^{x/12}$ and the height is also given which is $\sqrt{3}^{\frac{x}{6} }$ .

Area of triangle = $\frac{bh}{2}$ ,

substituting we will end up with 18 $\sqrt{3}$ =( 2 $\sqrt{3}^{x/12}$ * $\sqrt{3}^{\frac{x}{6} }$ ) / 2

Here is the tricky part $\sqrt{x} = x^{\frac{1}{2} }$ , i totally forgot about that lol.

Simplifying: we will get 18 $\sqrt{3}$ = $3^{\frac{x}{8} }$

Now, in order to find the x, we will need to take the log of both sides.

$log18\sqrt{3} = \frac{x}{8} log3$

Solving for x we end up getting:

$8\frac{log18\sqrt{3} }{log3}$ = x

where x = 25.047.

To be honest I deserve a nobel prize not a brainliest lol.

Good question bro, take it easy.

5 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

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

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

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

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

Kareem and Marco went to the mall. Kareem spent $8 less than Marco spent at the mall. Kareem spent more than $15. Write an inequ

mestny [16]

Marco spent more than $23. M > 23

We know the Kareem spent 8 less than Marco and that Kareem spent more than 15. If we add 8 onto the 15, we know that Marco has to be more than 23.

To graph this, draw an open circle at 23 and have an arrow going to the right.

4 0

3 years ago

Evaluate 3(a-4b) when a= -2 and b= 5

pshichka [43]

I believe the answer is -26 :) I hope this helps! ^_^

7 0

3 years ago

ALGEBRA QUESTION x+7=8.3;x=1.3 TRUE OR FALSE