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Zigmanuir [339]

3 years ago

9

For every 15 situps Ryan does, he does 10 push-ups. Create a table of equivalent ratios that show the relationship between the n

umber of sit-ups and the number of push-ups Ryan does.

1 answer:

goblinko [34]3 years ago

8 0

Answer:

15/10, 30/20, 60/40, 120/80, 240/160, 480/320

Step-by-step explanation:

The number of push-ups is 2/3 the amount of sit-ups, so the ratio of sit-ups to push-ups is 3:2.

The easy way to do this is to double both 15 and 10 so your answer will stay an equivalent ratio. You will always have a 3:2 ratio this way.

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2 years ago

Standardized exam's scores are normally distributed. In a recent year, the mean test score was 21.2 and the standard deviation

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3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

"A study conducted at a certain college shows that 56% of the school's graduates find a job in their chosen field within a year

KiRa [710]

Answer:

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they find a job in their chosen field within one year of graduating, or they do not. The probability of a student finding a job in their chosen field within one year of graduating is independent of other students. 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.

56% of the school's graduates find a job in their chosen field within a year after graduation.

This means that $p = 0.56$

Find the probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

This is $P(X \geq 1)$ when n = 6.

Either none find a job, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{6,0}.(0.56)^{0}.(0.44)^{6} = 0.0073$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0073 = 0.9927$

99.27% probability that among 6 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating.

8 0

3 years ago

Enter the exponents for the variables of the GCF of the following monomials.

miskamm [114]

The smallest exponents of the x and y term are 2. The Greatest Common Factor ( G C F ) for 9 x² y² and 5 x² y³ is: x² y².

8 0

3 years ago

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