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earnstyle [38]
3 years ago
15

Please help ASAP!! please no thank you have a great day!

Mathematics
2 answers:
Archy [21]3 years ago
7 0

1 and 5 are yes and the rest are no

Delvig [45]3 years ago
6 0

Answer: See explanation

Step-by-step explanation:

67 × 100 = 6700

<em>67 × (10 x 10)      </em><em>Yes</em>

10 × 10 = 100

67 × 100 = 6700

<em>67 × (10 × 10 × 10)      </em><em>No</em>

10 × 10 × 10 = 1000

67 × 1000 = 67000

<em>670 × 10³           </em><em>No</em>

10³ = 1000

670 × 1000 = 670000

<em>670        </em><em>No</em>

670 ≠ 6700

<em>6700      </em><em>Yes</em>

6700 = 6700

<em>67000      </em><em>No</em>

67000 ≠ 6700

Hope this helped!

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The coefficient of these two number are -24 and 7. :)
4 0
3 years ago
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Harold wants to get at a grade of 70 to 75 in his math class. His grade will be the average
ratelena [41]

Answer:

132

cause you need to multiple by one thousand to get your ans2er

3 0
2 years ago
Which values are outliers? 6. 1, 3. 9, 6. 4, 10. 6, 4. 4, 4. 5, 1. 6, 11. 2, 0. 8, 3. 9, 6. 5 Select Outlier or Not Outlier for
Kisachek [45]

Answer:

<em>10 and 11</em>

Step-by-step explanation:

  1. An outlier is a number that is much larger or smaller than the general populous of numbers. For example, if there was a dot plot with dots clustered at 2, 3, and 5, but there was one or two dots around 10, then 10 would be considered an outlier.
  2. Let's put the numbers in order to see easier which numbers are outliers. I will not copy down repeating numbers.
  3. 0, 1, 2, 3, 4, 5, 6, 10, 11.
  4. 10 and 11 are isolated.

If I am incorrect in my reasoning, please let me know so that I can plan better for my future answers. Have an amazing day.

8 0
2 years ago
How to solve this problem x +9=-12
STALIN [3.7K]
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3 0
2 years ago
At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37
monitta

Answer:

Half-life of the goo is 49.5 minutes

G(t)= 300e^{-0.014t}

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let M_0\,,\,M_f denotes initial and final mass.

M_0=300\,\,grams\,,\,M_f=37.5\,\,grams

According to exponential decay,

\ln \left ( \frac{M_f}{M_0} \right )=-kt

Here, t denotes time and k denotes decay constant.

\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014

So, half-life of the goo in minutes is calculated as follows:

\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes

Half-life of the goo is 49.5 minutes

\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}

So,

G(t)= M_f=M_0e^{-kt}

Put M_0=300\,\,grams\,,\,k=0.014

G(t)= 300e^{-0.014t}

Put t = 32 minutes

G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams

7 0
3 years ago
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