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3 years ago

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Find (5.2 x 107)(5 x 108) (3.2 x 106)(5 x 105) , expressed in scientific notation. A) 1.625 x 104 B) 1.625 x 105 C) 8.125 x 104

D) 8.125 x 105

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

I think it is A) 1.625 x 10^4

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Malik deposited 1050 in a savings account and it earned 241.50 in simple interest after four years. Find the interest after four

kap26 [50]

Answer:

if it's true simple interest then each year's interest is the same so interest for one year is a quarter of 241.50 which give you the rate but if each year's interest in left in the account the result will be different , it is a really poorly worded question

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3 years ago

Find the l.c.m of x²+x, x²-1, x²-x

olga2289 [7]

Answer:

X³-x

Step-by-step explanation:

Lcm

x(x-1)(x+1)

x(x²-1)

x³-x

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2 years ago

A triangle has sides with lengths 5, 12, and 13. Verify this is a Pythagorean triple. Approximate the acute angles in this trian

djyliett [7]

A triangle with lengths 5, 12, and 13 is a Pythagorean triple

The approximate acute angles are 22.6° and 67.4°

Pythagorean triple

Pythagorean triple consist of positive number(a, b, c) such that it obeys the rule:

c² = a² + b²

A triangle whose sides form a Pythagorean triple is called a right angle triangle.

The longest side is the hypotenuse side.

Therefore,

5² + 12²
25 + 144 = 169
√169 = 13

Therefore,

5² + 12² = 13²

This means the triangle is a Pythagorean triple.

<h3>Acute angles</h3>

Acute angles are angles that are less than 90 degrees. This means the other two angles are acute angle.

Therefore, let's find them

tan ∅ = opposite / adjacent = tan ∅ = 5 / 12 = ∅ = 22.6198649155 = 22.6°
180 - 22.6 - 90 = 67.4°

learn more on Pythagorean triple here: brainly.com/question/2293263?referrer=searchResults

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2 years ago

Read 2 more answers

Factor y 2 + 10z - 10y - yz.

dedylja [7]

Y² + 10z - 10y - yz = (y - 10)(y - z)

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3 years ago

34% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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

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

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 $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0

3 years ago