Write the following decimal number in it’s equivalent fraction form. Show all work for full credit. -0.6

Molodets [167]

8.67 (rounded to the nearest tenth)

6 0

3 years ago

Casey has 48 peppers and 16 yellow peppers . If he wants to put the peppers into baskets so that each basket has the same number

liraira [26]

Answer:

3 red peppers and 1 yellow pepper in each basket.

Step-by-step explanation:

We have been given that Casey has 48 peppers and 16 yellow peppers. He wants to put the peppers into baskets so that each basket has the same number of red and yellow peppers. We are asked to find the number of red and yellow papers in each basket.

To solve our given problem, we need to find greatest common factor of 48 and 16.

Prime factorization of 48 is $2\times 2\times 2\times 2\times 3$ .

Prime factorization of 16 is $2\times 2\times 2\times 2$ .

Greatest common factor of 48 and 16 would be $2\times 2\times 2\times 2=16$ .

Therefore, Casey can make 16 baskets with same number of red and yellow peppers.

Now we will divide 48 and 16 by 16 to find number of red peppers and yellow peppers in each basket respectively.

$\text{Red peppers in each baseket}=\frac{48}{16}=3$

$\text{Red peppers in each baseket}=\frac{16}{16}=1$

Therefore, there will be 3 red peppers and 1 yellow pepper in each basket.

6 0

4 years ago

In a recent study on world happiness, participants were asked to evaluate their current lives on a scale from 0 to 10, where 0

uysha [10]

Answer:

a) A response of 8.9 represents the 92nd percentile.

b) A response of 6.6 represents the 62nd percentile.

c) A response of 4.4 represents the first quartile.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.9

Standard Deviation, σ = 2.2

We assume that the distribution of response is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) We have to find the value of x such that the probability is 0.92

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.92$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 1.405\\x = 8.991 \approx 8.9$

A response of 8.9 represents the 92nd percentile.

b) We have to find the value of x such that the probability is 0.62

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.62$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = 0.305\\x = 6.571 \approx 6.6$

A response of 6.6 represents the 62nd percentile.

c) We have to find the value of x such that the probability is 0.25

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 5.9}{2.2})=0.25$

Calculation the value from standard normal z table, we have,

$P(z$

$\displaystyle\frac{x - 5.9}{2.2} = -0.674\\x = 4.4172 \approx 4.4$

A response of 4.4 represents the first quartile.

4 0

4 years ago

A varies directly as b, and a =4 when b=24 write an equation that relates a and b

uysha [10]

Y=kx
K=y/x

SO:
A=kb
K=a/b

5 0

3 years ago

Data on the modulus of elasticity obtained 1 minute after loading in a certain configuration and 4 weeks after loading for the s

mixer [17]

Answer:

The Confidence Interval = (2180, 2782)

Lower Bound for the true average difference between 1-minute modulus and 4-week modulus = 2180

Upper Bound for the true average difference between 1-minute modulus and 4-week modulus = 2782

- The interval obtained represents the range of values that the true average difference between 1-minute modulus and 4-week modulus can take on.

Step-by-step explanation:

The confidence interval for any distribution is the range of values that the true population mean can take on with some level of confidence.

It is given mathematically as

Confidence Interval = (Sample Mean) ± (Margin of Error)

The Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as

Margin of Error = (Critical Value) × (Standard Error of the Mean)

Before anything, we first state the conditions necessary for the confidence interval calculated to be valid.

- The sample must be a random sample extracted from the population distribution using random sampling technique.

- The sample distribution must be normal or approximately normal and this is confirmed by knowing that the population distribution is normal or almost normal.

These two conditions are assumed to be plausible for this sample distribution.

Here, we need the confidence interval for the true average difference between 1-minute modulus and 4-week modulus.

So, before anything, we need to first fin the sample mean and standard deviation for the differences obtained from the sample specimens.

904, 3370, 2580, 3117, 2850, 2690, 2180, 1241, 2210, 2350, 2260, 2758, 2880, 2750, 3520, 2033

Mean = (Σx/N)

Σx = Sum of all the variables = 39693

N = Sample size = 16

Sample Mean = (39693/16) = 2480.8125

Standard deviation = $\sqrt{\frac{Σ(x - xbar)^{2}}{N-1}\\}$

x = each variable

xbar = sample mean = 2480.8125

Note that we use (N-1) for the standard deviation because it is a sample standard deviation.

$\sqrt{\frac{Σ(x - xbar)^{2}}{N-1}\\}$ = √(7227548.4375/15) = √(481836.5625)

= 694.14448243863 = 694.1445

Confidence Interval = (Sample Mean) ± (Margin of Error)

Margin of Error = (Critical Value) × (Standard Error of the Mean)

Confidence Interval = (Sample Mean) ± [(Critical Value) × (Standard Error of the Mean)]

Critical Value is obtained from the t-distribution table since no information on the population standard deviation is known.

For that, we need the significance level for the test and the degree of freedom.

Significance level = α = 0.05 (given in the question)

Degree of freedom = N - 1 = 16 - 1 = 15

t(0.05, 15) = 1.753 (from the t-distribution tables)

Standard error of the Mean = σₓ = (σ/ $\sqrt{N}$ ) = (694.1445/ $\sqrt{16}$ ) = 173.536125 = 173.536

Confidence Interval = (Sample Mean) ± [(Critical Value) × (Standard Error of the Mean)]

CI = (2480.8125) ± [(1.753 × 173.536)]

CI = (2480.8125) ± (301.09385)

CI = (2179.71865, 2781.90635)

CI = (2180, 2782)

This interval represents the range of values that the true average difference between 1-minute modulus and 4-week modulus can take on.

Hope this Helps!!!

3 0

3 years ago