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aivan3 [116]
4 years ago
12

Round 584,565 to the nearest hundred thousand.

Mathematics
2 answers:
lora16 [44]4 years ago
8 0

Answer:

600,000

Step-by-step explanation:

600,000

600,000

600,000

600,000

WINSTONCH [101]4 years ago
7 0

Answer:

600000

Step-by-step explanation:

5<u>8</u>4565

The underlined number is 8. Is that bigger or smaller than 5?

BIGGER!!

So go to the hundred thousands place and add one to 5. That eqauls 6 so the answer is 600000

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(3 - 8a)(-1)<br> please help asap
katrin2010 [14]

Answer:

-3 + 8a

Step-by-step explanation:

multiply 3 by -1 as well as -8a by -1 and you'll get -3 + 8a

6 0
3 years ago
Read 2 more answers
In 2002, the temperature on the first day of summer was 79°F. This is 4°F less than it was on the first day of summer the previo
kap26 [50]

Answer:

  C. In 2001, the temperature was 83°F.

Step-by-step explanation:

  79 is 4 less than 83.

The temperature in 2001 was 83 °F.

5 0
3 years ago
The sum of Pete's and Sam's ages is 30. Five years ago, Pete was 3 times as old as Sam. How old is Sam? Let P = Pete's age, S =
m_a_m_a [10]
Five years ago:
Peter= P-5
Sam = S-5
Peter was 3 times older than Sam, which means that (P-5) = 3(S-5)
P-5 = 3S-15
So, the equation that would complete the system is C. P - 5 = 3S - 15

P-5 = 3S-15
P-3S = -15+5
P-3S = -10

So, now we use two equations: P-3S = -10 and the original one P+S=30
P-3S=-10
P+S=30 (to simplify the equation, multiply it by -1)

P-3S=-10
-P-S=-30
__________
0P-4S=-40
-4S=-40
4S=40
S=40/4
S=10

P+S=30
P+10=30
P=20

This means that Pete is 20, and Sam is 10. Hope this helps!



4 0
3 years ago
Read 2 more answers
5. Let f be a continuous and differentiable function for all real numbers. The only critical points for f are located
ollegr [7]
Ok so the answer is super simple, its A
6 0
3 years ago
Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=10
Tju [1.3M]

Answer: 0.8238

Step-by-step explanation:

Given : Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with \mu=106 and \sigma=15.

Let x denotes the scores on a certain intelligence test for children between ages 13 and 15 years.

Then, the proportion of children aged 13 to 15 years old have scores on this test above 92 will be :-

P(x>92)=1-P(x\leq92)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{92-106}{15})\\\\=1-P(z\leq })\\\\=1-P(z\leq-0.93)=1-(1-P(z\leq0.93))\ \ [\because\ P(Z\leq -z)=1-P(Z\leq z)]\\\\=P(z\leq0.93)=0.8238\ \ [\text{By using z-value table.}]

Hence, the proportion of children aged 13 to 15 years old have scores on this test above 92 = 0.8238

4 0
3 years ago
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