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3 years ago

Express loga 6 + loga 70 as a single logarithm

Mathematics

2 answers:

ICE Princess25 [194]3 years ago

8 0

Answer:

logₐ(420)

Step-by-step explanation:

Katena32 [7]3 years ago

4 0

Answer:

The answer is

$log_{a}(420)$

Step-by-step explanation:

You have to use Logarithm Law,

$log_{a}(b) + log_{a}(c) ⇒ log_{a}(b \times c)$

* Take note, number b and c can only be multiplied when they have the same base, a

So for this question :

$log_{a}(6) + log_{a}(70)$

$= log_{a}(6 \times 70)$

$= log_{a}(420)$

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0

3 years ago

You have been doing research for your statistics class on the prevalence of severe binge drinking among teens. You have decided

Serggg [28]

Answer:

(1.218 ; 1.322)

the confidence interval is appropriate

Step-by-step explanation:

The confidence interval :

Mean ± margin of error

Sample mean = 1.27

Sample standard deviation, s = 0.80

Sample size, n = 914

Since we are using tbe sample standard deviation, we use the T table ;

Margin of Error = Tcritical * s/√n

Tcritical at 95% ; df = 914 - 1 = 913

Tcritical(0.05, 913) = 1.96

Margin of Error = 1.96 * 0.80/√914 = 0.05186

Mean ± margin of error

1.27 ± 0.05186

Lower boundary = 1.27 - 0.05186 = 1.218

Upper boundary = 1.27 + 0.05186 = 1.322

(1.218 ; 1.322)

According to the central limit theorem, sample means will approach a normal distribution as the sample size increases. Hence, the confidence interval is valid, the sample size of 914 gave a critical value at 0.05 which is only marginally different from that will obtained using a normal distribution table. Hence, the confidence interval is appropriate

8 0

3 years ago

Answer the questions below. When you are finished, submit this assignment to your teacher by the due date for full credit. Total

Andrews [41]

Jill had a total score of 40 on 13 quizzes, so her average was
40/13 ≈ 3.08

_____
1 quiz @ 1 = total of 1
3 quizzes @ 2 = total of 6
4 quizzes @ 3 = total of 12
4 quizzes @ 4 = total of 16
1 quiz @ 5 = total of 5
total score = 1 + 6 + 12 + 16 + 5 = 40
total quizzes = 1 + 3 + 4 + 4 + 1 = 13

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3 years ago

Composition about myself in 150 words

tigry1 [53]

Answer:

I am a student who’s studying in a prestigious college in Bangalore city. It’s a city where I grew up. I live in a town with my family. The school I studied till 12th is also in the town.

Things I am good at

Almost everyone is kind at least one sport. The one competition that I am good at is basketball. In my school, almost everyone had an obsession with the sport, and so did I. Every game period, my teacher would make us play basketball, along with other games. Over the years, the way I played basketball improved, and while learning the game, I discovered other lessons as well. One of the lessons I’ve learned is how to play in a team. When you play in a group, you depend on each other for winning.

I have always been energetic and lively. While many people feel awkward and weird, making me friends, I have no problems with making new friends. I can talk to everyone quickly and know them.

This is not about me took it from the web!!

Hope it helps!!!

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3 years ago

This is for my sister

ELEN [110]

Answer:

Large umber is 11, small one 7

Step-by-step explanation:

11- 4 =7

7+11=18

5 0

3 years ago

Read 2 more answers