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Answer:
To make a profit of at least $500 she must sell at least 12 necklaces
Step-by-step explanation:
Joyce rented a booth at a carnival at a cost of $95 to sell handmade beaded necklaces. The cost of making and packaging each beaded necklace was $15. If Joyce sells the beaded necklaces at $35 each then we have to find beaded necklaces she sell to make a profit of at least $500.
The costs side of the equation, we have:
Let no. of necklaces that she sell are x
∴ The cost of making and packaging x beaded necklace is $15x
Total Cost = 95+15x
Now, Joyce sells the beaded necklaces at $35 each. Therefore, selling price will be $35x
To make a profit of at least $500 the equation can be written as
⇒
Hence, to make a profit of at least $500 she must sell at least 12 necklaces
Answer:
No, we can't reject the dealer's claim with a significance level of 0.05.
Step-by-step explanation:
We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.
You take a random sample of 600 cars in San Francisco and find that 141 are blue.
<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>
SO, Null Hypothesis, : p = 25% {means that 25% of all cars in San Francisco are blue}
Alternate Hypothesis, : p
25% {means that % of all cars in San Francisco who are blue is different from 25%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of 600 cars in San Francisco who are blue =
= 0.235
n = sample of cars = 600
So, <u><em>test statistics</em></u> =
= -0.866
<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>
Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.