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Alexxx [7]
2 years ago
6

Lincoln Middle School has a total of $200 to spend on notebooks and tablets for the school board meeting. Notebooks cost $7 each

, and tablets cost $5 each. If x represents the number of notebooks and y represents the number of tablets, which graph correctly shows the solution to the problem?
Mathematics
1 answer:
Anton [14]2 years ago
4 0
The school can buy fifteen notebooks and nineteen tablets.  15X7=105 and 5X19=95.  95+105=200.  The first thing you want to do is realize that 5Xanything will give you an answer that has a 0 or a five at the end, so you want to find a product by 7 that ends with a 0 or 5 too.
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If $10,000 is invested at x percent simple annual interest for n years, which of the following represents the total amount of in
Ganezh [65]

Answer: C. 10000n\dfrac{x}{100}

Step-by-step explanation:

Given : The principal amount : P=$10,000

Rate of interest (in percent ) : x

Time period :  n years

The formula to calculate the simple interest is given by :-

S.I.=\dfrac{P\times R\times T}{100}

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it can also be written as \Rightarrow\ S.I.=10000n\dfrac{x}{100}

Hence, C is the right option.

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Which table represents a linear function? A, B, C, D
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Answer:

Table B represents a linear function.

Step-by-step explanation:

Table B is a linear function because y goes up by a constant, steady rate of 4 every time x goes up by one.

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gtnhenbr [62]
We are given with
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yKpoI14uk [10]

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No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

H_0: \mu=2\\\\H_a:\mu< 2

The significance level is 0.01.

The sample has a size n=52.

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As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

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t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803

The degrees of freedom for this sample size are:

df=n-1=52-1=51

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

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