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3 years ago

1210^5 or 1.210^6 which one is closer to 1 million? please explain.

2 answers:

6 0

12*10^5 = 12*100000 = 1200000 ;
1200000 - 1000000 = 200000 ;
1.2*10^6 = 1.2*1000000 = 1200000;
1200000 - 1000000 = 200000;
Because 2000000 = 200000, neither of the two numbers is closer to 1 milion !

leonid [27]3 years ago

5 0

$12 \times 10^5 = 1.2 \times 10^6$ Therefore, they are both equally close to 1 million.

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Ben's broker charges a commission of $9.95 for every $1000 of stock bought or sold.

irga5000 [103]

Answer:

Ben made $4,908.96 profit

up 31.9%

Step-by-step explanation:

Not really sure how the broker is figuring his commission. Some round to the nearest whole $1000, others pro-rate as I have done below.

Ben bought for 600($15.67) + $9.98(600(15.67)/1000) = $9,495.83

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3 years ago

8 packs of 12 units is equaled to

GREYUIT [131]

Not to sure but wouldn't be 96

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4 years ago

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Label From least to greatest 1/2, 1/3, 1/,6, 4/9

Sholpan [36]

Answer:

1/6 4/9 1/3 1/2

Step-by-step explanation:

3 0

4 years ago

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal

faust18 [17]

Answer:

The probability is $P(X \le 70) = 0.9685$

Step-by-step explanation:

From the question we are told that

The mean values is $M_1 = 15 \ minutes , \ M_2 = 30\ minutes, \ M_3 = 20 \ minutes$

The standard deviation is $\sigma_1 = 1 \ minutes , \ \sigma_2 = 2 \ minutes , \ \sigma = 1.5 \ minutes$

Generally the total mean is mathematically represented as

$\= x = \sum M_i$

=> $\= x = 15 + 30 + 20$

=> $\= x = 65$

Generally the total variance is mathematically represented as

$V(x) = \sum \sigma^2_i$

=> $V(x) = 2^2 + 1^2 + 1.5^2$

=> $V(x) = 7.25$

Generally the total standard deviation is mathematically represented as

$\sigma = \sqrt{V(x)}$

=> $\sigma = \sqrt{7.25 }$

=> $\sigma = 2.69$

Generally the probability that it takes at most 70 min of machining time to produce a randomly selected component is mathematically represented as

$P(X \le 70) = 1 - P( X > 70 )$

Here $P(X > 70) = P(\frac{X - \mu }{\sigma} > \frac{70 - 65}{ 2.69} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

$P(X > 70) = P(Z > 1.8587 )$

From the z table

The area under the normal curve to the right corresponding to 1.8587 is

$P(Z > 1.8587 ) = 0.031535$

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=> $P(X \le 70) = 0.9685$

7 0

3 years ago

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3 years ago

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12*10^5 or 1.2*10^6 which one is closer to 1 million? please explain.

1210^5 or 1.210^6 which one is closer to 1 million? please explain.