Michaela signs up for a package of 6 dance classes for $29.95, and each additional dance class is $3.95.

olive's garden in only 10y2 and the watermelon plants she wants to grow required 2.5y2 each.how many watermelon plants can she g

DerKrebs [107]

Answer:

4 plants

Step-by-step explanation:

In this problem, we are given the total area of the garden, which is:

$A=10y^2$

The number of plants that can fit inside the garden is given by the equation:

$A=na$

where:

n is the number of plants

a is the area of each plants

In this problem, the area of each plant is

$a=2.5y^2$

Therefore, we can re-arrange the equation to find the number of plants that can fit:

$n=\frac{A}{a}=\frac{10}{2.5}=4$

8 0

3 years ago

Suppose it is known that 8 out of the 20 teams in the sample had a season winning percentage better than 0.500.

Ganezh [65]

Answer:

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

8 out of the 20 teams in the sample had a season winning percentage better than 0.500. This means that $n = 20, \pi = \frac{8}{20} = 0.4$ .

95% confidence interval

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.96\sqrt{\frac{0.6*0.4}{20}} = 0.1853$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.96\sqrt{\frac{0.6*0.4}{20}} = 0.6147$

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).

3 0

4 years ago

5.13570<br><img src="https://tex.z-dn.net/?f=5.1%20%5Cdiv%2035.7" id="TexFormula1" title="5.1 \div 35.7" alt="5.1 \div 35.7" ali

anzhelika [568]

0.1428 is your answer

7 0

4 years ago

Let Bold u comma Bold v comma Bold w 1 comma Bold w 2 comma and Bold w 3 be vectors in Bold Upper R Superscript n. If u and v ca

Sonja [21]

Answer:

Step-by-step explanation:

given that U, V are two vectors in R^n

These two vectors can be written as a linear combination of 3 vectors

w1, w2, and w3

To prove that U+V also can be written as a linear combination of these three vectors.

Since U is a linear combination we can write for not all a,b, c equal to 0

$U = aw1+bw2+cw3$

Similarly for d,e,f not all equal to 0

$V= cw1+dw2+ew3$

Adding these we have

$U+V =(a+d)w1 + (b+e) w2+(c+f)w3$

Here all a+d, b+e or c+f cannot be simultaneously 0.

So we get U+V can be written as a linear combination of w1, w2 w3 as follows:

$U+W = gw1+hw2+iw3 \\g = a+d\\h = b+e\\i = c+f$

Proved

4 0

3 years ago

Order the value of the quotient from the smallest (top) to largest (bottom)

Vsevolod [243]

Answer:

6/11 divided by 2 2/3 (=9/44)

2/7 divided by 4/5 (=5/14 or 15/42 scaled up)

1 6/7 divided by 3 (= 13/21 or 26/42 scaled up)

3/10 divided by 1/8 (=2 2/5)

Step-by-step explanation:

You cannot divide a fraction by a fraction. Instead, multiply by the reciprocal ( flip it)

2/7 divided by 4/5= 2/7 x 5/4

Multiply straight across the top, 2x5= 10

10 is the numerator

Now multiply straight across the bottom.

7x4= 28

28 is the denominator.

10/28 is the answer to this problem, however it can be simplified. Both 10 and 28 are divisible by 2. This simplifies to 5/14.

For mixed numbers, you first have to make them improper fractions.

Ex: 2 2/3

Take the denominator (3) multiply it by the whole number (2) and add the numerator (2)

3x2=6 (now add the remaining 2 from the numerator) 6+2= 8. 8 is your new numerator and you will keep the existing denominator of 3 making the improper fraction 8/3.

8 0

3 years ago