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Darya [45]
3 years ago
15

kenzie purchased three DVD and one music CD.The CD cost kenzie $12.99. If he paid the same amount for each DVD and spent 42.96,h

ow much did each DVD cost.
Mathematics
1 answer:
lidiya [134]3 years ago
3 0
42.96-12.99= 29.97/3= 9.99
Take the total cost subtract it by the cost of the CD then divide the remainder by 3.

Kenzie spent $9.99 on each DVD
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A recipe take 90 minutes to make.25% of the time is spent preparing the ingredients.How many minutes are spent preparing
joja [24]
It will take 22.5 min cause 25 divided by 100 wuld be 4 wich then divide 90 with
4 0
2 years ago
When Victoria goes bowling, her scores are normally distributed with a mean of 130
Artemon [7]

Answer:

0.2377 or about a 23.77% chance

Step-by-step explanation:

P(123<X<130) = normalcdf(123,130,130,11) = 0.2377303466 ≈ 0.2377

Therefore, the probability that the next game Victoria bowls, her score will be  between 123 and 130 is 0.2377 or about a 23.77% chance

5 0
3 years ago
How many solutions does the equation 6(x -2) = 8 + 6x have?
Contact [7]

Answer:

There are no solutions.

Step-by-step explanation:

First simplify the equation.

<u>6(x - 2)</u> = 8 + 6x

Distribute.

6x - 12 = 8 + 6x

Adding 8 to a number and subtracting 12 from the same number to get the same answer doesn't make sense so there are no solutions.

7 0
2 years ago
In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s
Oksi-84 [34.3K]

Answer:

a) \mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

b) From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

c) P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

P(Z\geq2.070)=1-P(Z

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

\mu_{\bar x} =\mu = 276.1

\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3

Part b

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)

Part c

For this case we want this probability:

P(\bar X \geq 285)

And we can use the z score defined as:

z=\frac{\bar x -\mu}{\sigma_{\bar x}}

And using this we got:

P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

And using a calculator, excel or the normal standard table we have that:

P(Z\geq2.070)=1-P(Z

8 0
3 years ago
Please help me with my homework idk if it is right or wrong
jeka57 [31]
Q1: 7×6 = 42
Q2: 42×1/8 = 42/8 = 5.25 (5 1/4) in^3
Q3: (0.5)^3 = 0.5×0.5×0.5 = 0.125 m^3
Q4: 4(5/2)^2 = 4×5/2×5/2 = 4×25/4 = 25 in^3
Q5: 3(13/2)(3/2) = 117/4 = 29.25 (29 1/4) in^3
Q6: (0.9)^3 = 0.729 cm^3
6 0
3 years ago
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