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Lina20 [59]
3 years ago
14

Kala bought two types of cheese at a deli. She bought 0.50 pound of American cheese and 1.25 pounds of Swiss cheese. The bar dia

gram and equation below represent Kala’s purchase, where p represents the total number of pounds of cheese she bought.
What is the total number of pounds of cheese Kala bought?
0.75
1.20
1.30
1.75
Mathematics
2 answers:
ElenaW [278]3 years ago
7 0
The correct answer is 1.75 because the equation is p=.50+1.25 which equals 1.75. 
AleksAgata [21]3 years ago
6 0

Answer:

The total number of pounds of cheese Kala bought is 1.75 pounds

Step-by-step explanation:

Kala bought two types of cheese at a deli.

She bought 0.50 pound of American cheese and 1.25 pounds of Swiss cheese.

The total number of pounds of cheese Kala bought

= 0.50+1.25

= 1.75 pounds

Hence,  the total number of pounds of cheese Kala bought is:

  1.75 pounds

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ivanzaharov [21]
First let’s try to cancel out the x
5x + -5x = 0
Add the y and the numbers together
-3y + -2y = -5y
26 + -16 = 10
-5y = 10
y = -2
Use y=-2 in one of the equations
-3(-2) + 5x = 26
6 + 5x = 26
5x = 20
X = 4

So
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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

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Answer:

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