Which of the following are more precise than a weight of 1,000 pounds? (Check all that apply)

What is the solution of the equation (y – 6) = 2y - 3?

schepotkina [342]

Answer:

y=-3

Step-by-step explanation:

7 0

2 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

$Z = \frac{X - \mu}{s}$

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

$Z = \frac{X - \mu}{s}$

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago

Find the height of a parallelogram with an area of 56 and a base of 20.

qaws [65]

The correct answer is 2.8

6 0

3 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

2 years ago

If she can pick 52 apples in one hour how many can she pick in 45 min

inn [45]

Answer:

39 apples

Step-by-step explanation:

45 is 3/4 or 75% of an hour. So, multiply 52 by 75% to find the answer.

75% ÷ 100 = 0.75

52 x 0.75 = 39

5 0

2 years ago

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