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Bad White [126]

1 year ago

14

Two students are using estimation to determine reasonable solutions to the expression 89.1 times 9.3. Katie uses the expression

90 times 10. Amaya uses the expression 89 times 9. Which is the best comparison of the estimates and the actual product?

2 answers:

dmitriy555 [2]1 year ago

5 0

Amaya’s because she rounded to the nearest tenth correctly if the number behind the decimal had been 5 or above Katie would be right but because the number is 4 or below you round down therefore 89.1 would be 89 and 9.3 would be 9

Lady bird [3.3K]1 year ago

5 0

Answer:

D. Katie’s estimate will be greater than the actual product, and Amaya’s estimate will be less than the actual product.

Step-by-step explanation:

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Sergio has p paintings in his art collection. He and other local painting collectors agreed to donate a total of 48

Tom [10]

Answer:

P - 4

Step-by-step explanation:

Sergio and the other painting collectors have decided to donate a total of 48 paintings all together.

We know that there are 12 collectors in total and they will each donate the same number of paintings.

48 paintings in total

12 collectors

48 / 12 = 4

We now know Sergio will donate 4 paintings from his collection of P paintings. Sergio will have P - 4 paintings left.

We do not know what "P" is equal to, so we cannot give an exact number for how many paintings he will have left. However, we know he will have 4 fewer paintings after donating.

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A monthly service fee of $0.50 and a per-check fee of $0.50.

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2 years ago

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A member of a student team playing an interactive marketing game received the fol- lowing computer output when studying the rela

nirvana33 [79]

Answer:

$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case for the significance level assumed $\alpha=0.05$ we see that $p_v >\alpha$ so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

Step-by-step explanation:

Let's suppose that we have the following linear model:

$y= \beta_o +\beta_1 X$

Where Y is the dependent variable and X the independent variable. $\beta_0$ represent the intercept and $\beta_1$ the slope.

In order to estimate the coefficients $\beta_0 ,\beta_1$ we can use least squares procedure.

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: $\beta_1 = 0$

Alternative hypothesis: $\beta_1 \neq 0$

Or in other words we want to check is our slope is significant (X have an effect in the Y variable )

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal

The significance level assumed on this case is $\alpha=0.05$

The standard error for the slope is given by this formula:

$SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}$

Th degrees of freedom for a linear regression is given by $df=n-2$ since we need to estimate the value for the slope and the intercept.

In order to test the hypothesis the statistic is given by:

$t=\frac{\hat \beta_1}{SE_{\beta_1}}$

The p value on this case would be given by:

$p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91$

So on this case for the significance level assumed $\alpha=0.05$ we see that $p_v >\alpha$ so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

3 0

2 years ago

A jet flies 425 km from Ottawa to Québec at rate v + 60. On the return flight, the

Marina CMI [18]

Answer:

a. $\frac{- 42,500}{(v + 60)(v - 40)}$

Step-by-step Explanation:

Given:

Distance Ottawa to Québec = 425 km

Initial flight rate = v + 60

Return flight rate = v - 40

$t = \frac{d}{r}$

Required:

Flight times difference of the initial and return flights

Solution:

=>Flight time of the initial flight:

$t = \frac{d}{r}$

$t = \frac{425}{v + 60}$

=>Flight time of the return flight:

$t = \frac{425}{v - 40}$

=>Difference in flight times:

$\frac{425}{v + 60} - \frac{425}{v - 40}$

$\frac{425(v - 40) -425(v + 60)}{(v + 60)(v - 40)}$

$\frac{425(v) - 425(40) -425(v) -425(+60)}{(v + 60)(v - 40)}$

$\frac{425v - 17000 -425v - 25500}{(v + 60)(v - 40)}$

$\frac{425v - 425v - 17000 - 25500}{(v + 60)(v - 40)}$

$\frac{- 42,500}{(v + 60)(v - 40)}$

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The answer for the first question: C The rotated image may be larger than the original image.

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