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2 years ago

Solve the following logarithmic equations. log2(49x^2) = 4

Mathematics

1 answer:

Mariana [72]2 years ago

4 0

Answer:

$x_{1} = \frac{4}{7} \\x_{2} = \frac{-4}{7} \\$

Step-by-step explanation:

To solve this equation, one must apply the following logarithmic property:

$log_{a}(b) = c\\$

then

$b= a^{c}$

Applying it to the problem at hand:

$2^{4} = 49x^{2} \\x^{2} = \frac{16}{49} \\x_{1} = \frac{4}{7} \\x_{2} = \frac{-4}{7} \\$

The solutions to the problem are 4/7 and -4/7.

*Note that this solution was pretty straight forward because log2(16) = 4 is a known value, otherwise, a change of base to a base ten log would be required.

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Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp

yKpoI14uk [10]

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

$H_0: \mu=2\\\\H_a:\mu< 2$

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803$

The degrees of freedom for this sample size are:

$df=n-1=52-1=51$

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

3 0

2 years ago

3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses.

Rufina [12.5K]

Answer:

a) The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 , P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

8 0

2 years ago

A company with $60,000 in sales last year had $69,000 in sales this year. Assuming the company's sales continue to grow at the s

aalyn [17]

Subtract 69000 from 60000. Answer becomes 9000. So you add 9000 to 69000 2 times. Your answer becomes 87000.

5 0

2 years ago

Read 2 more answers

Determine if this conjecture is true. If not, give a counterexample. The difference of two negative numbers is a negative number

Orlov [11]

That would be a false statement

6 0

2 years ago

How many numbers from 10 to 99 have a tens place exactly 3 times greater than their ones place? PLZZZ answer this question . I w

patriot [66]

Answer:

Step-by-step explanation:

2 digit number starts from 10 ends at 99

between 10 and 19 there is only one number whose tens digit is more than ones digit.

that is 10

between 20 and 29 there are two numbers

20 and 21

like the same

between 30 and 39 there are 3 numbers

10–19. 1

20–29. 2

30–39. 3

40–49. 4

50–59. 5

60–69. 6

70–79. 7

80–89. 8

99–99. 9

sum of first n natural numbers is n(n+1)/2

9(9+1)/2=45

4 0

2 years ago