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3 years ago

Round 593 to nearest ten Also are you in 2nd grade cause I am in 3rd

Mathematics

2 answers:

Tanzania [10]3 years ago

5 0

Answer:

590

Step-by-step explanation:

To the ten that is nearest to 593

gladu [14]3 years ago

5 0

Answer:

590

Step-by-step explanation:

Because your ones place is not higher than 5 the tens place stays the same and everything before becomes zero

You might be interested in

What answer is this?

klemol [59]

Answer:

option B

$\frac{31}{50}-\frac{4}{25} i$

Step-by-step explanation:

Given in the question a complex fraction

To divide complex numbers, you must multiply by the conjugate. To find the conjugate of a complex number all you have to do is change the sign between the two terms in the denominator.

$\frac{5+4i}{6+8i}(\frac{6-8i}{6-8i})$

Distribute (or FOIL) in both the numerator and denominator to remove the parenthesis.

$\frac{30+24i-40i-32i²}{36-64i^{2} }$

Simplify the powers of i, specifically remember that i² = –1.

$\frac{62-16i}{36+64}$

$\frac{62}{100}-\frac{16i}{100}$

simply

$\frac{31}{50}-\frac{4}{25}i$

8 0

3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

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 combinations 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.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

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

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

2 years ago

A car rental agency has 150 cars. The owner finds that at a price of $48 per day, he can rent all the cars. For each $2 increase

hram777 [196]

Given that for each <span>$2 increase in price, the demand is less and 4 fewer cars are rented.

Let x be the number of $2 increases in price, then the revenue from renting cars is given by
$(48 + 2x) \times (150 - 4x)=7,200+108x-8x^2$ .

Also, given that f</span><span>or each car that is rented, there are routine maintenance costs of $5 per day, then the total cost of renting cars is given by
$5(150-4x)=750-20x$

Profit is given by revenue - cost.
Thus, the profit from renting cars is given by
</span><span> $(7,200+108x-8x^2)-(750-20x)=6,450+128x-8x^2$

For maximum profit, the differentiation of the profit function equals zero.
i.e.
</span><span> $\frac{d}{dx} (6,450+128x-8x^2)=0 \\ \\ 128-16x=0 \\ \\ x= \frac{128}{16} =8$

The price of renting a car is given by 48 + 2x = 48 + 2(8) = 48 + 16 = 64.

Therefore, the </span><span>rental charge will maximize profit is $64.</span>

3 0

3 years ago

G(x)=-x^2-1-2x<br> f(x)= x+5<br> Find (g-f)(x)

sertanlavr [38]

Djdhhggfgcsdddhgdfgccfgfdgxstgdrg

4 0

3 years ago

More Algebra hw-- :/

grin007 [14]

Answer:
-11x=-66
x=6

Explanation:
-11x-12=-78
1) Add 12 to both sides.
-11x=-78+12
-11x=-66
2) Divide both sides by -11.
x=-66/-11
x=6

Hope this helps! Please leave a thanks and brainliest answer if you can :)

3 0

2 years ago