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3 years ago

Which of the following are more precise than a weight of 10 pounds?

2 answers:

5 0

I guess any ten pounds can be the same weight as light stuffs combined together such as multiple breaks, a school bag, etc.

aleksley [76]3 years ago

4 0

well 30 pounds i think i hope im not wrong

You might be interested in

The volume of a cone is 62.8 cubic inches. The height of the cone is 15 inches. What is the radius of the cone, rounded to the n

Setler79 [48]

R = sqrt 3 * (V /( pi * h))
V = 62.8
pi = 3.14
h = 15
now we sub
R = sqrt 3 * (62.8 / (3.14 * 15)
R = sqrt 3 * (62.6 / 47.1)
R = sqrt 3 * 1.33
R = sqrt 3.99
R = 1.9 rounds to 2 inches <===

3 0

3 years ago

Read 2 more answers

Can someone help me solve these two problems? i need to find what x is.

Bond [772]

Answer:

1. x=2

2. x= 0.33

Step-by-step explanation:

1.So to start off you have a coefficient for 2 so you have to find what 2 to the power of 2 is =4 so then you have your new problem 4x+5=13 next you need to get 4x by itself so

-5 -5

then you have 4x=8. you divide by 4 on both sides

4. 4

so then you get x=2

2.So same here find what 3 to the power of 2 is =9

New equation 4x-1=9x+4x-4. Then combine like terms on the right you cant do it on the left because that is a separate part and those dont merge like terms so

9x+4x = 13x

new equation. 4x-1=13x-4

next subtract 4x on both sides

13x-4x=9x

new equation. -1=9x-4

next add 4 to both sides

-1+4= 3

new equation. 3=9x

so then divide 9 on both sides

3÷9= 0.33

x= 0.33

Hope this made sense and helps

7 0

3 years ago

A roster for a baseball team has 12 players. the coach is randomly making the batting order for 9 players. what is the probabili

andrey2020 [161]

The probability of you being the leadoff hitter is 8.3%.

Explanation:

The number of possible rosters can be calculated with permutations, knowing that the coach can choose the first hitter among all 12 players, then the second hitter among the 11 players remaining and so on until the 9th hitter among the remaining 4 players.

Therefore:
Possible rosters = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 = 79833600

The leadoff hitter is the first player in the batting order, therefore the number of possible rosters in which you are chosen as the first one is:
You leadoff = 1 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 = 6652800

The probability of you being leadoff is:
P = You leadoff / Possible rosters
 = 6652800 / 79833600
 = 1 / 12
 = 0.08333
 = 8.3%

Note that this is exactly the probability of you being chosen out of the 12 players of the team in a one-pick choice because it does not matter how the rest of the batting order is composed.

3 0

3 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

If Z > 2, the result is considered significantly high.

If $np \geq 10$ and $n(1-p) \geq 10$ , the binomial distribution can be approximated to the normal with:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

In this problem:

943 peas, thus, $n = 943$
3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$ .

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$ , which is 1 subtracted by the p-value of Z when X = 716.5.