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Oduvanchick [21]

4 years ago

12

Paul wants to build this model with clay but he does not know how many cubic centimeters of play to purchase how much play shoul

d he purchased

1 answer:

WITCHER [35]4 years ago

7 0

There isn't a picture so I can't see the model to help you.

You might be interested in

1. The value of (301)^2 – (300)^2 is

Ganezh [65]

Answer:
(301)^2-(300)^2
it is in the form of
a^2-b^2=(a+b)(a-b)
so,
(301+300)(301-300)
(601)(1)=601

7 0

3 years ago

The cost of renting a car for a day is $0.50 per mile plus a $15 flat fee.

Vadim26 [7]

Answer: a) y=0.50x+15

b) The graph of this equation form on a coordinate plane is a line.

c) Slope =0.50 and y-intercept = 15

Step-by-step explanation:

Let x = Number of miles driven by car

Given: The cost of renting a car for a day is $0.50 per mile plus a $15 flat fee.

a) Total cost = 0.50x+15

If y =total cost of renting the car, then y=0.50x+15 (i)

b) Above equation is similar to y= mx+c (ii) [m = slope , xc=y-intercept] which a linear equation .

So the graph of this equation form on a coordinate plane is a line.

c) Comparing (i) and (ii)

m=0.50 , c=15

Slope =0.50 and y-intercept = 15

3 0

3 years ago

In a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a

djverab [1.8K]

Answer:

90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].

Step-by-step explanation:

We are given that a a random sample of 10 residents of the state of Washington, the mean waste recycled per person per day was 2.3 pounds with a standard deviation of 0.39 pounds.

Firstly, the pivotal quantity for 90% confidence interval for the mean waste recycled per person per day for the population of Washington is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waste recycled per person per day = 2.3 pounds

s = sample standard deviation = 0.39 pounds

n = sample of residents = 10

$\mu$ = population mean waste recycled per person per day

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.833 < $t_9$ < 1.833) = 0.90 {<u>As the critical value of t at 9 degree of</u>

<u>freedom are -1.833 & 1.833 with P = 5%</u>}

P(-1.833 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.833) = 0.90

P( $-1.833 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.833 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.833 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ , $2.3-1.833 \times {\frac{0.39}{\sqrt{10} } }$ ]

= [2.074 , 2.526]

Therefore, 90% confidence interval for the mean waste recycled per person per day for the population of Washington is [2.074 , 2.526].

8 0

3 years ago

Every Thursday, Matt and Dave's Video Venture has “roll-the-dice" day. A customer may choose to roll two fair dice and rent a se

shepuryov [24]

Using the normal distribution, it is found that there is a 0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.

In a <em>normal distribution</em> 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.

For n instances of a normal variable, the mean is $n\mu$ and the standard error is $s = \sigma\sqrt{n}$

In this problem:

Mean of $0.47, standard deviation $0.15, hence $\mu = 0.47, \sigma = 0.15$

30 instances, hence $n\mu = 30(0.47) = 14.1, s = 0.15\sqrt{30} = 0.8216$

The probability is <u>1 subtracted by the p-value of Z when X = 15</u>, hence:

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

Considering the n instances:

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

$Z = \frac{15 - 14.1}{0.8216}$

$Z = 1.1$

$Z = 1.1$ has a p-value of 0.8643.

1 - 0.8643 = 0.1357.

0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.

A similar problem is given at brainly.com/question/25769446

6 0

2 years ago

How much is 75% of 12 liters of milk

Tresset [83]

The awnser for this is 9 I think it sure

7 0

3 years ago