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3 years ago

Please help! need to know how to solve this Urgent!

Mathematics

2 answers:

Artemon [7]3 years ago

6 0

= $log_{ 3}({x}^{2} - 4) - log_{3}(x - 2) + log_{3}(x)$

= $log_{ 3}(\dfrac{{x}^{2} - 4}{x - 2}) + 2 × 1$

= $log_{ 3}(\dfrac{(x-2)(x+2)}{x - 2}) + 2 × 1$

= $log_{ 3}(x+2) + 2 × 1$

PilotLPTM [1.2K]3 years ago

5 0

Step-by-step explanation:

I dont know if this helps but that what I came up with

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Write and solve a system of equations for this question please

andre [41]

Let b and s be the velocity of the boat and stream respectively....

3(b-s)=27

b-s=9

b=9+s we will use this in the second equation....

2(b+s)=30

b+s=15, now use b found above in this to get:

9+s+s=15

2s+9=15

2s=6

s=3, and since from earlier b-s=9, b=9+s so

b=12

So the speed of the boat is 12mph and the stream is 3 mph

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Answer:

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Step-by-step explanation:

Substitute the values into the equation

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<h2><em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>

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3 years ago

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3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0

3 years ago