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3 years ago

Racehorse weighs 470 kilograms. How much does it weigh in grams

Mathematics

2 answers:

tigry1 [53]3 years ago

7 0

Answer: 470000 grams

Step-by-step explanation:

1 kilogram =1000 grams

470×1000=470000

Wewaii [24]3 years ago

5 0

Answer:

it weighs 470000 grams ...

Step-by-step explanation:

1 kg = 1000

470 kg = 470×1000 => 470000

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How can 4.17 be written as a whole number with a fraction?

solmaris [256]

It can be written as 4 17/100

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3 years ago

Whoever answers I’ll mark brainliest

Andre45 [30]

Answer:

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2 years ago

A phone company offers two monthly plans. Plan A costs $10 plus an additional $0.15 for each minute of calls. Plan B costs $30 p

Mkey [24]

First, you need to write to expressions to model each situation:

Plan A: 10+0.15x

Plan B: 30+0.1x

Next, set the expressions equal to each other and solve for x:

10+0.15x=30+0.1x

*Subtract 0.1x from both sides to isolate the variable*

10+0.05x=30

*Subtract 10 from both sides*

0.05x=20

*Divide both sides by 0.05*

x=400

The plans would have the same cost after 400 minutes of calls.

To find how much money the plans cost at 400 minutes, plug 400 into either expression. We'll use Plan A:

10+0.15(400)

10+60

The plans will cost $70.

Hope this helps!

3 0

3 years ago

Can someone explain the process getting from:

spin [16.1K]

$\rm 5=e^{3b}$

The unknown b is stuck in the exponent position.
We can can fix that by using logarithms.
Log is the inverse operation of the exponential.

We'll take log of each side.
Log of what base tho?

Well, the base of our exponential is e,
so we'll take log base e of each side.

$\rm log_e(5)=log_e(e^{3b})$

We'll apply one of our log rules next:
$\rm \log(x^y)=y\cdot\log(x)$

This allows us to take the exponent out of the log,

$\rm log_e(5)=(3b)log_e(e)$

Another thing to remember about logs:
When the base of the log matches the inside of the log,
then the whole thing is simply 1,
$\rm log_{10}(10)=1$
$\rm log_5(5)=1$
$\rm log_e(e)=1$

So our equation simplifies to this,

$\rm log_e(5)=(3b)\cdot1$

As a final step, divide both sides by 3,

$\rm \frac13log_e(5)=b$

k, hope that helps!

6 0

3 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

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3 years ago