All categories

4 years ago

Solve log x2 + 1 = 5. Round to the nearest thousandth if necessary.

Mathematics

2 answers:

Lesechka [4]4 years ago

6 0

Answer:

the solutions are x = -100,100

Step-by-step explanation:

The given logarithmic equation is $\log x^2 +1=5$

Subtract 1 to both sides

$\log x^2=4$

Remove logarithm, we get

$x^2=10^4$

Take square root both sides

$\sqrt{x^2}=\pm\sqrt{10^4}\\\\x=\pm10^2\\\\x=\pm100$

Therefore, the solutions are x = -100,100

aleksley [76]4 years ago

5 0

Answer:

x = 100

Step-by-step explanation:

You might be interested in

Need help ASAP with number 5

vredina [299]

Costs would be nine dollars for the lemons and four dollars for the sugar ($-13.00). You end up losing only fifty cents throughout, so you would have had to sell (.5×25=12.50) 25 cups to pay for all but fifty cents.

Answer is 25 cups

4 0

3 years ago

A number to the 8th power divided by the same number to the 5th power equals 27. What is the number?

Yuliya22 [10]

x^8/x^5=27

x=3 when you flip the equation (rounded number)

7 0

3 years ago

Read 2 more answers

Which model is most appropriate for the data shown in the graph below?

Klio2033 [76]

The answer is Quadratic

5 0

4 years ago

Read 2 more answers

Thank you for your help>>>>>>>>>>+++++++

Shkiper50 [21]

Answer:

Step-by-step explanation:

5 0

3 years ago

A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o

ryzh [129]

Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑ $_{x=0$ b( x, n, p )

= ³∑ $_{x=0$ b( x, 10, 0.6 )

= ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.6)^x$ $( 1 - 0.6 )^{10-x$

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.3 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.4 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.5 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

3 0

3 years ago