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Naddika [18.5K]

2 years ago

10

Of the flowers on Jill's front lawn,2/5 are tulips. Of the tulips,5/8 are yellow. What fraction of the flowers on Jill's front l

awn are yellow tulips. A 18 B 1/4 C 1/2 D 7/13

2 answers:

Anit [1.1K]2 years ago

5 0

2/5 of the flowers are tulips, 5/8 of the tulips are yellow. Yellow tulpis are 5/8 of 2/5 of the flowers.

$\frac{5}{8} \ of \ \frac{2}{5}=\frac{5}{8} \times \frac{2}{5}=\frac{1}{8} \times \frac{2}{1}=\frac{2}{8}=\boxed{\frac{1}{4}} \Leftarrow \hbox{answer B}$

Andrew [12]2 years ago

3 0

Answer:

1/4

Step-by-step explanation:

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2 years ago

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The average time taken to complete an exam, X, follows a normal probability distribution with mean = 60 minutes and standard dev

k0ka [10]

Answer: b. 0.8413

Step-by-step explanation:

Given : The average time taken to complete an exam, X, follows a normal probability distribution with $\mu=60\text{ minutes}$ and $\sigma=30\text{ minutes}$ .

Then, the probability that a randomly chosen student will take more than 30 minutes to complete the exam will be :-

$P(x>30)=P(z>\dfrac{30-60}{30})\ \ [\because\ z=\dfrac{x-\mu}{\sigma} ]\\\\=P(z>-1)=P(z-z)=P(Z$

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2 years ago

A bag holds 12 red marbles, 11 green marbles, 17 blue marbles, and 5 yellow marbles. What is the probability that you will not c

Nataly_w [17]

Hey there! Welcome to Brainly!

Let's see how many marbles we have in total.

12+11+17+5=45

We want to find the probability of selecting a marble that is not blue. Let's see how many marbles aren't blue.

12+11+5=28

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28/48

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7 0

2 years ago

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The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction

Len [333]

Answer:

(a) <em> </em><em>n</em> : 20 50 100 500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of <em>n</em> = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of <em>n</em> = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of <em>n</em> = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of <em>n</em> = 500

$P(\mu-200$

$=P(-1.86$

<em> n</em> : 20 50 100 500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0

2 years ago

Report Error Suppose $P(x)$ is a polynomial of smallest possible degree such that: $\bullet$ $P(x)$ has rational coefficients $\

motikmotik

Answer:

We want a polynomial of smallest degree with rational coefficients with zeros in $\sqrt{7}$ , $1 - \sqrt{6}$ and -3. The last root gives us the factor (x+3). Hence, our polynomial is

$P(x) =(x+3)q(x)$

where $q$ is a polynomial with rational coefficients and roots $\sqrt{7}$ and $1 - \sqrt{6}$ . The root $\sqrt{7}$ gives us a factor $x-\sqrt{7}$ , but in order to obtain rational coefficients we must consider the factor $x^2-7$ .

An analogue idea works with $1 - \sqrt{6}$ . For convenience write $x - 1 + \sqrt{6} = ( x - 1) + \sqrt{6}$ . This gives the factor $(x-1)^2-6$ . Hence,

$P(x) = (x+3)(x^2-7)((x-1)^2-6)=x^5+x^4-18x^3-22x^2+77x+105$

Notice that $P(-1)=24$ . So, in order to satisfy the last condition we divide by 3 the whole polynomial, without altering its roots. Finally, the wanted polynomial is

$P(x) =(1/3)x^5+(1/3)x^4-6x^3-(22/3)x^2+(77/3)x+35$

Step-by-step explanation:

We must have present that any polynomial it's determined by its roots up to a constant factor. But here we have irrational ones, in order to eliminate the irrational coefficients that a factor of the type $x-\sqrt7$ will introduce in the expression, we need to multiply by its conjugate $x+\sqrt7$ . Hence, we will obtain $x^2-7$ that have rational coefficients. Finally, the last condition is given with the intention to fix the constant factor. Usually it is enough to evaluate in the point and obtain the necessary factor.

4 0

2 years ago