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3 years ago

Fifteen is what percent of 1500

Mathematics

2 answers:

marysya [2.9K]3 years ago

6 0

Divide 15 by 1500 and get your percentage. 1% is the answer

Simora [160]3 years ago

3 0

The answer is 225

Solution:

1500 move one decimal point

It is going to be 150 then divide it by 2 it is going to be 75 then add 150 then the answer is going to be 225

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What are the solutions to 0=x^2-x+1?

kumpel [21]

Answer: No solutions

Step-by-step explanation:

There are no solutions to 0 = x² - x + 1.

[Method One]

We can solve this by rewriting the equation, but it is quicker to graph it when there are no solutions. See attached.

[Method Two]

I said we can solve this by solving an equation, here it is. We will use the quadratic formula since the equation given is already in the proper form.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

$\displaystyle x=\frac{1\pm\sqrt{(-1)^2-4(1)(1)} }{2(1)}$

$\displaystyle x=\frac{1\pm\sqrt{-3} }{2}$

-> Negative numbers don't have real square roots, so there is no solution

4 0

2 years ago

1. When professional golfer Jordan Spieth hits his driver, the distance the ball travels can be modeled by

Doss [256]

Answer:

a. About 96% of Spieth's drives travel at least 290 yards

b. The ball has to travel about 314.24 yards

Step-by-step explanation:

* Let us explain how to solve the problem

- The distance the ball travels can be modeled by distribution with

mean 304 yards and standard deviation 8 yards

- Jordan would need to hit the ball at least 290 yards to have a clear

second shot that avoids a large group of trees

- We need to find the percent of Spieth's drives travel at least

290 yards

∵ The mean μ = 304 yards

∵ The standard deviation σ = 8 yards

∵ The distance x = 290

∵ P(x ≥ 290) = P(z ≥ z)

- We need to find z score

∵ z-score = (x - μ)/σ

∴ z = $\frac{290-304}{8}$ = -1.75

* Let us use the normal distribution table to find the corresponding

area of z = -1.75

∵ P(-z) = 1 - P(z)

∵ P(z ≥ -1.75) = 1 - 0.04006 = 0.95994

∴ P(x ≥ 290) = 0.96 = 96%

* <em>About 96% of Spieth's drives travel at least 290 yards</em>

- On another golf hole, Spieth has the opportunity to drive the ball

onto the green if he hits the ball a distance in the top 10% of all

his drives

- We need to find how far the ball has to travel

- Let us find the z-score from the normal distribution table for the 10%

to the right or 90% to the left

∵ The area which equivalent to 0.9 ≅ 0.89973

∴ z = 1.28

∵ z-score = (x - μ)/σ

∴ 1.28 = $\frac{x-304}{8}$

- Multiply both sides by 8

∴ 10.24 = x - 304

- Add 304 for both sides

∴ x = 314.24 yards

* <em>The ball has to travel about 314.24 yards</em>

8 0

3 years ago

Which label would NOT be appropriate for volume.

Dafna1 [17]

Answer:

Please provide more information so the question could be answered

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

When the product of 6 and the square of a number is increased by 5 times the number, the result is 4. Which equation represents

adelina 88 [10]

$6{x}^{2} + 5x = 4$
$6 {x}^{2} + 5x - 4 = 0$

7 0

3 years ago

Read 2 more answers

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Fifteen is what percent of 1500​

Fifteen is what percent of 1500