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3 years ago

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Question 1 WRONG With what number must 3.475817 be multiplied in order to obtain the number 34,758.17? A 100 1,000 10,000 D 100,

000 Question 1 Explanation: The correct answer is (C). The decimal must be moved four places to the right. To do this, we must multiply by a number with four zeroes. The correct answer is 10,000. Question 2 How much greater is the value of 3x+5 than the value of 3x−7?

1 answer:

lbvjy [14]3 years ago

5 0

Answer:

1 - Option C - 10,000

2 - 3x+5 will be 12 more than 3x-7.

Step-by-step explanation:

1) To find : With what number must 3.475817 be multiplied in order to obtain the number 34,758.17?

Solution :

The number 3.475817 is multiplied by number to obtain 34,758.17.

As we see that, the decimal is moved four places to the right.

Which means the number is multiplied by 10000.

i.e. $3.475817\times 10000=34,758.17$

Therefore, option C is correct.

2) To find : How much greater is the value of 3x+5 than the value of 3x−7?

Solution ;

Assuming that x is the same value in each equation,

In order to determine the difference between the two values,

We ignore the 3x part as it is the same on both of the equations.

So, from 5 to -7 i.e. 7+5=12.

Therefore, 3x+5 will be 12 more than 3x-7.

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2 years ago

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#15(A)
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2 years ago

Need answer to 4 and 5

algol13

Answer:

Substitute n =1, 2, 3, 4, 5 to get the first five terms.

Step-by-step explanation:

(4). $g_n = 2 . 3^{n - 1}$

To find the first term, substitute n = 1.

Therefore, $g_1 = 2 . 3^{1 - 1} = 2 . 3^{0} = 2 . 1$ = 2

$g_2 = 2 . 3^{2 - 1} = 2 . 3$ = 6

$g_3 = 2 . 3^{3 - 1} = 2 . 3^{2} = 2 . 9$ = 18

$g_4 = 2 . 3^{4 - 1} = 2 . 3^{3} = 2 . 27$ = 54

$g_5 = 2 . 3^{5 - 1} = 2 . 3^{4} = 2 . 81$ = 168

Therefore the first five terms of the sequence are: 2, 6, 18, 54, 168.

(5). $t_n = \frac{2}{3}t_{n - 1}$

$t_2 = \frac{2}{3} t_{2 - 1} = \frac{2}{3}t_1 = \frac{2}{3}6$ = 4

$t_3 = \frac{2}{3} t_2 = \frac{2}{3}. 4 = \frac{8}{3}$

$t_4 = \frac{2}{3}t_3 = \frac{2}{3}\frac{8}{3} = \frac{16}{9}$

$t_5 = \frac{2}{3}t_4 = \frac{2}{3}\frac{16}{9} = \frac{32}{27}$

Therefore the first five terms in the sequence are: 6, 4, 8/3, 16/9, 32/27.

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3 years ago

The weight distribution of parcels sent in a certain manner is normal with mean value 17 lb and standard deviation 3.3 lb. The p

Yakvenalex [24]

Answer: The value of c would be 26.514 lb.

Step-by-step explanation:

Since we have given that

Mean = 17 lb

Standard deviation = 3.3 lb

At 99% level of significance, z = 2.58

So, it becomes,

$Z=\dfrac{X-\mu}{\sigma}\\\\2.58=\dfrac{X-17}{3.3}\\\\2.58\times 3.3=X-17\\\\8.514+17=X\\\\X=25.514$

So, the weight c would be

$c=25.514+1\\\\c=26.514$

Hence, the value of c would be 26.514 lb.

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3 years ago

If a baseball player get a hit in 31% of his at-bats, what is the probability that he will

solong [7]

Given:

A baseball player get a hit in 31% of his at-bats.

To find:

The probability that he will get a hit 5 at-bats in a row.

Solution:

Since, a baseball player get a hit in 31% of his at-bats, therefore

Probability that he get a hit = $\dfrac{31}{100}$

= $0.31$

Now, the probability that he will get a hit 5 at-bats in a row is

$P=0.31\times 0.31\times 0.31\times 0.31\times 0.31$

$P=(0.31)^5$

$P=0.0028629151
$

Therefore, the required probability is $(0.31)^5$ and it can be written as 0.0028629151 .

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2 years ago