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3 years ago

Could you solve it? I don't know how to solve it:(

Mathematics

1 answer:

Nikitich [7]3 years ago

5 0

Answer:

-1, 4

Step-by-step explanation:

$log_2x+ log_2(x-3)=2 \\ log_2 \{x(x-3) \}=2 \\ x(x - 3) = {2}^{2} \\ {x}^{2} - 3x = 4 \\ {x}^{2} - 3x - 4 = 0 \\ {x}^{2} - 4x + x - 4 = 0 \\ x(x - 4) + 1(x - 4) = 0 \\ (x - 4)(x + 1) = 0 \\ x - 4 = 0 \: \: or \: \: x + 1 = 0 \\ x = 4 \: \: or \: \: x = - 1 \\ \huge \red{ \boxed{x = \{ - 1, \: \: 4 \}}}$

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Kate gave 5/6 of her CDs to Sue. Sue gave 7/8 of these to Joan. Joan gave 3/5 of these to Ann. Ann received 70 CDs how many did

olya-2409 [2.1K]

Let us assume the number of CDs Kate had = x
Number of CDs given to Sue = 5x/6
Number of CDs given to Joan = (5x/6) * ( 7/8)
= 35x/48
Number of CDs given to Ann = (35x/48) * (3/5)
= 7x/16
Total number of CDs Ann received = 70
Then
7x/16 = 70
x = 160
From the above deduction, it can be concluded that the number of CDs that Kate originally had was 160. I hope that this is the answer that has actually come to your help.

5 0

3 years ago

The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the

Degger [83]

Using the normal distribution, we have that:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem, the parameters are given as follows:

$\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358$

Hence:

The distribution of X is $X \approx (57,22)$ .

The distribution of $\mathbf{\bar{X}}$ is $\bar{X} \approx (57, 5.3358)$ .

The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:

X = 58:

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

$Z = \frac{58 - 57}{22}$

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

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

$Z = \frac{55 - 57}{22}$

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

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

$Z = \frac{58 - 57}{5.3358}$

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

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

$Z = \frac{55 - 57}{5.3358}$

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0

2 years ago

Determining a Local Maximum and Minimum

g100num [7]

Answer: