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kompoz [17]
3 years ago
12

An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situati

on using money as the context. Then, evaluate the expression and interpret the result in terms of the situation. Here's an example.
The following situation is modeled by the expression -24.50 + 100.00: Jonathan overdrew his checking account by $24.50. He then transferred $100.00 from his savings account to his checking account. His new checking account total is $75.50.

Make sure to use complete sentences in your answers.
1) $8.15 + $11.99
2) $20 - $18.04
3) 7($1.45)
4) 42.08 / 4
Mathematics
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

Question

An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situation using money as the context. Then, evaluate the expression and interpret the result in terms of the situation. Here's an example.

The following situation is modeled by the expression -24.50 + 100.00: Jonathan overdrew his checking account by $24.50. He then transferred $100.00 from his savings account to his checking account. His new checking account total is $75.50.

Make sure to use complete sentences in your answers.

1) $8.15 + $11.99

2) $20 - $18.04

3) 7($1.45)

4) 42.08 / 4

It's too short. Write at least 20 characters to explain it well.

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If 1 cup of milk is added to a 3-cup mixture that is 2/5 flour and 3/5 milk, what percent of the 4-cup mixture of milk?
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mariarad [96]

Answer:

p_v =P(z>2.087) =0.0184

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

\sigma=2.5

And the statistic is given by this formula:

z=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}

Where z follows a normal standard distribution

The system of hypothesis on this case are:

Null hypothesis: \mu_c \leq \mu_t+1

Alternative hypothesis: \mu_c >\mu_t+1

Or equivalently:

Null hypothesis: \mu_c - \mu_t \leq 1

Alternative hypothesis: \mu_c-\mu_t>1

Our notation on this case :

n_c =45 represent the sample size for group control

n_t =45 represent the sample size for group  treatment

\bar X_c =5.2 represent the sample mean for the group control

\bar X_t =3.1 represent the sample mean for the group treatment

And now we can calculate the statistic:

z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087  

And now we can calculate the p value using the alternative hypothesis:

p_v =P(z>2.087) =0.0184

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

5 0
3 years ago
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