Asked by the big ones
Add 7.3×107+1.8×107 . Which is the correct sum?(1 point)
Responses
9.1×107
9.1 times 10 superscript 7 baseline
8.1×107
8.1 times 10 superscript 7 baseline
9.1×1014
9.1 times 10 superscript 14 baseline
9.1+107
Responses
9.1×107
9.1 times 10 superscript 7 baseline
8.1×107
8.1 times 10 superscript 7 baseline
9.1×1014
9.1 times 10 superscript 14 baseline
9.1+107
Answers
Answered by
the big ones
The population of a city is given as 3.45×106 . A nearby city has a population of 1.3×105 . What is the sum of the populations of the two cities in scientific notation?(1 point)
Responses
4.75×105
4.75 times 10 superscript 5 baseline
4.75×1011
4.75 times 10 superscript 11 baseline
35.8+105
35.8 plus 10 superscript 5 baseline
3.58×106
Responses
4.75×105
4.75 times 10 superscript 5 baseline
4.75×1011
4.75 times 10 superscript 11 baseline
35.8+105
35.8 plus 10 superscript 5 baseline
3.58×106
Answered by
the big ones
An empty dump truck weighs about 2.8×105 pounds. How much does it weigh if 15,300 pounds of trash is added? The answer should be in scientific notation.(1 point)
Responses
4.33×105 pounds
4.33 times 10 superscript 5 baseline pounds
15,302.8 pounds
15,302.8 pounds
2.953×105 pounds
2.953 times 10 superscript 5 baseline pounds
29,530 pounds
Responses
4.33×105 pounds
4.33 times 10 superscript 5 baseline pounds
15,302.8 pounds
15,302.8 pounds
2.953×105 pounds
2.953 times 10 superscript 5 baseline pounds
29,530 pounds
Answered by
the big ones
The half-life of uranium-235 is 700,000,000 years. The half-life of uranium-238 is 4.463×1011 years longer. What is the half-life of uranium-238? The answer should be in decimal form.(1 point)
Responses
516,300,000,000
516,300,000,000
11,463,000,000
11,463,000,000
447,000,000,000
447,000,000,000
4.4637×1011
Responses
516,300,000,000
516,300,000,000
11,463,000,000
11,463,000,000
447,000,000,000
447,000,000,000
4.4637×1011
Answered by
the big ones
A mobile phone holds 1.28×1011 bytes of storage. You can also pay more money to get an additional 384,000,000,000 bytes. How much storage would your phone have if you buy the additional bytes? The answer should be in scientific notation.(1 point)
Responses
385.28×1011
385.28 times 10 superscript 11 baseline
385,280,000,000
385,280,000,000
512×109
512 times 10 superscript 9 baseline
5.12×1011
Responses
385.28×1011
385.28 times 10 superscript 11 baseline
385,280,000,000
385,280,000,000
512×109
512 times 10 superscript 9 baseline
5.12×1011
Answered by
GPT-4o mini
To add \( 7.3 \times 10^7 \) and \( 1.8 \times 10^7 \), you can add the coefficients (the numbers in front) while keeping the exponent the same:
\[
7.3 + 1.8 = 9.1
\]
So the sum is:
\[
9.1 \times 10^7
\]
Therefore, the correct response is:
**9.1×10^7** (9.1 times 10 superscript 7 baseline)
\[
7.3 + 1.8 = 9.1
\]
So the sum is:
\[
9.1 \times 10^7
\]
Therefore, the correct response is:
**9.1×10^7** (9.1 times 10 superscript 7 baseline)
Answered by
GPT-4o mini
To find the sum of the populations of the two cities:
1. The populations are given as:
- City A: \( 3.45 \times 10^6 \)
- City B: \( 1.3 \times 10^5 \)
2. Before adding, it's often easier to express both numbers in terms of the same power of ten. Let’s convert City B's population to the same exponent as City A:
\[
1.3 \times 10^5 = 0.13 \times 10^6
\]
3. Now we can add the populations:
\[
3.45 \times 10^6 + 0.13 \times 10^6 = (3.45 + 0.13) \times 10^6 = 3.58 \times 10^6
\]
Thus, the sum of the populations of the two cities in scientific notation is:
**3.58×10^6** (3.58 times 10 superscript 6 baseline)
1. The populations are given as:
- City A: \( 3.45 \times 10^6 \)
- City B: \( 1.3 \times 10^5 \)
2. Before adding, it's often easier to express both numbers in terms of the same power of ten. Let’s convert City B's population to the same exponent as City A:
\[
1.3 \times 10^5 = 0.13 \times 10^6
\]
3. Now we can add the populations:
\[
3.45 \times 10^6 + 0.13 \times 10^6 = (3.45 + 0.13) \times 10^6 = 3.58 \times 10^6
\]
Thus, the sum of the populations of the two cities in scientific notation is:
**3.58×10^6** (3.58 times 10 superscript 6 baseline)
Answered by
GPT-4o mini
To find the total weight of the dump truck after adding 15,300 pounds of trash, we can follow these steps:
1. The weight of the empty dump truck is given as \( 2.8 \times 10^5 \) pounds.
2. First, we need to express 15,300 pounds in scientific notation.
\[
15,300 = 1.53 \times 10^4
\]
3. Now, we will need to add \( 2.8 \times 10^5 \) (the weight of the truck) to \( 1.53 \times 10^4 \) (the weight of the trash). To do this, we should express both numbers in terms of the same power of ten. We can convert \( 1.53 \times 10^4 \) to \( 10^5 \):
\[
1.53 \times 10^4 = 0.153 \times 10^5
\]
4. Now we can add the two weights:
\[
2.8 \times 10^5 + 0.153 \times 10^5 = (2.8 + 0.153) \times 10^5 = 2.953 \times 10^5
\]
The total weight of the dump truck with the added trash in scientific notation is:
**2.953×10^5 pounds** (2.953 times 10 superscript 5 baseline pounds)
1. The weight of the empty dump truck is given as \( 2.8 \times 10^5 \) pounds.
2. First, we need to express 15,300 pounds in scientific notation.
\[
15,300 = 1.53 \times 10^4
\]
3. Now, we will need to add \( 2.8 \times 10^5 \) (the weight of the truck) to \( 1.53 \times 10^4 \) (the weight of the trash). To do this, we should express both numbers in terms of the same power of ten. We can convert \( 1.53 \times 10^4 \) to \( 10^5 \):
\[
1.53 \times 10^4 = 0.153 \times 10^5
\]
4. Now we can add the two weights:
\[
2.8 \times 10^5 + 0.153 \times 10^5 = (2.8 + 0.153) \times 10^5 = 2.953 \times 10^5
\]
The total weight of the dump truck with the added trash in scientific notation is:
**2.953×10^5 pounds** (2.953 times 10 superscript 5 baseline pounds)
Answered by
GPT-4o mini
To find the half-life of uranium-238, we start with the half-life of uranium-235 and add the additional time given.
1. The half-life of uranium-235 is \( 700,000,000 \) years.
2. The half-life of uranium-238 is \( 4.463 \times 10^{11} \) years longer than that of uranium-235.
First, we need to convert \( 4.463 \times 10^{11} \) to decimal form:
\[
4.463 \times 10^{11} = 446,300,000,000
\]
Now, we can compute the half-life of uranium-238:
\[
\text{Half-life of uranium-238} = 700,000,000 + 446,300,000,000
\]
To do the addition:
\[
700,000,000 + 446,300,000,000 = 447,000,000,000
\]
Thus, the half-life of uranium-238 is:
**447,000,000,000** years.
1. The half-life of uranium-235 is \( 700,000,000 \) years.
2. The half-life of uranium-238 is \( 4.463 \times 10^{11} \) years longer than that of uranium-235.
First, we need to convert \( 4.463 \times 10^{11} \) to decimal form:
\[
4.463 \times 10^{11} = 446,300,000,000
\]
Now, we can compute the half-life of uranium-238:
\[
\text{Half-life of uranium-238} = 700,000,000 + 446,300,000,000
\]
To do the addition:
\[
700,000,000 + 446,300,000,000 = 447,000,000,000
\]
Thus, the half-life of uranium-238 is:
**447,000,000,000** years.
Answered by
GPT-4o mini
To find the total storage of the mobile phone after adding the additional bytes, we start with the storage capacity of the phone and the additional storage amount:
1. The storage of the mobile phone is given as \( 1.28 \times 10^{11} \) bytes.
2. The additional storage is given as \( 384,000,000,000 \) bytes. We can express this in scientific notation:
\[
384,000,000,000 = 3.84 \times 10^{11}
\]
Now, we can add the two capacities:
\[
1.28 \times 10^{11} + 3.84 \times 10^{11} = (1.28 + 3.84) \times 10^{11} = 5.12 \times 10^{11}
\]
Therefore, the total storage of the phone after purchasing the additional bytes is:
**5.12×10^11** (5.12 times 10 superscript 11 baseline) bytes.
1. The storage of the mobile phone is given as \( 1.28 \times 10^{11} \) bytes.
2. The additional storage is given as \( 384,000,000,000 \) bytes. We can express this in scientific notation:
\[
384,000,000,000 = 3.84 \times 10^{11}
\]
Now, we can add the two capacities:
\[
1.28 \times 10^{11} + 3.84 \times 10^{11} = (1.28 + 3.84) \times 10^{11} = 5.12 \times 10^{11}
\]
Therefore, the total storage of the phone after purchasing the additional bytes is:
**5.12×10^11** (5.12 times 10 superscript 11 baseline) bytes.
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