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4 years ago

Add. 4.8 ⋅ 106 5.2 ⋅ 107 express your answer in scientific notation.

Mathematics

1 answer:

andreev551 [17]4 years ago

4 0

(4.8 x 106) + (5.2 x 107) = 508.8 + 556.4 = 1,065.2 = 1.0652 x 10^3

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Factor 2n(n + 3) - 5(n + 3).

Vilka [71]

2n(n + 3) - 5(n + 3) = (2n-5)(n+3)

A. (n + 3) (2n - 5)

6 0

4 years ago

Ryobi Limited sold a set of saws to Ace Hardware. The list price was $3,800. Ryobi offered a chain discount of 8/3/1. What's the

laila [671]

Answer: Option 'B' is correct.

Our net price will be $3357.21.

Explanation:

Since we have given that

The list price of set of saws = $3800

There are 3 successive discounts :

8%,3%,1%

So, if we apply first for 8% discount our new price becomes,

$3800\times \frac{100-8}{100}=3800\times 0.92=\$3496$

Now, if we apply 3% on our deducted price , our new price becomes ,

$3496\times \frac{100-3}{100}=3496\times 0.97=\$3391.12$

Now, we'll apply for 1% on our new deducted price , our net price becomes,

$3391.12\times \frac{100-1}{100}=3391.12\times 0.99=\$3357.21$

Hence, our net price will be $3357.21.

Therefore, Option 'B' is correct.

6 0

3 years ago

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago

How many miles does a runner travel in a five-kilometer race?

ollegr [7]

Since one kilometer is equal to about 0.621 miles, a runner travels 3.105 miles in a five-kilometer race.

4 0

3 years ago

Solve for x: x/-5+6≤2

Anon25 [30]

Answer:

x ≥ 20

Step-by-step explanation:

hope this helps!

7 0

3 years ago

Read 2 more answers