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igor_vitrenko [27]

2 years ago

11

Natalie works in a shop and earn $43 per day. She earns an extra $3 for each toy she sells. If Natalie wants to earn at least $7

0 per day, which ineaquality shows the minimum number of toys, n, that she should sell?

2 answers:

vladimir2022 [97]2 years ago

8 0

70-43=27
27 divided by 3=9

She needs to sell 9 toys

Hope this helps!

Pachacha [2.7K]2 years ago

7 0

<span>This would be your answer 43 + 3n ≥ 70, so n ≥ 9

I hape this help</span>

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Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2\neq 0$

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$n_2 =5$ represent the sample size for group 2

We can calculate the mean and deviation for eaach group with the following formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

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$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

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$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}$

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$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

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