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2 years ago

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Heidi's mom made flower arrangements for a party. She made 4 times as many rose arrangements as tulip arrangements. Heidi's mom

made 40 arrangements. How many flower arrangements of each type did Heidi's mom make? Complete the bar model. write an equation and solve

2 answers:

djyliett [7]2 years ago

7 0

8 tulip arrangements
32 rose arrangments

juin [17]2 years ago

6 0

Answer:

<em>Heidi'd mom made 32 rose arrangements and 8 tulip arrangements.</em>

Step-by-step explanation:

Let the number of rose arrangements be = x

Let the number of tulip arrangements be = y

Heidi's mom made 4 times as many rose arrangements as tulip arrangements.

We get the equation;

$x=4y$

Heidi's mom made 40 arrangements. We get the second equation:

$x+y=40$

Substituting the value of x from first equation to second equation, we get

$4y+y=40$

=> $5y=40$

Dividing both sides by 5;

y = 8

And x = 4y

So, $x=4(8)=32$

x = 32

<em>Hence, Heidi'd mom made 32 rose arrangements and 8 tulip arrangements.</em>

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Answer:

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Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

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$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

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$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

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