If you combine the partial crates, how many additional full crates of oranges can be sold?

2 answers:

5 0

If you combine the partial crates, pending on how full and how many you have it will vary how many full crates of oranges can be sold.

Allisa [31]2 years ago

3 0

I think there is something that tells us how full the crates are, otherwise there's no way to solve this problem.
Let's assume that 5x oranges can fit into each crate, and that we have a crate with x oranges, a crate with 2x oranges, and a crate with 4x oranges. We get this:
x/5x+2x/5x+4x/5x
Add those together and we get
7x/5x
Turn it into a mixed number:
1 2x/5x
And we get our answer

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Determine whether the number could be the probability of an event. Explain your reasoning. -0.25

mr Goodwill [35]

yo sup??

The number can't be a probability of an event cause probability of an even lied between 0 and 1 whereas the number given to use is -0.25.

Since probability can't be negative therefore it isn't a probability of an event.

Hope this helps.

8 0

3 years ago

What is the surface area of this rectangular prism that has 9 ft a width of 5 feet and a height of 7.5ft

daser333 [38]

To get your answer:

9 times 5 times two= 90
9 times 7.5 times two= 135
5 times 7.5 times two= 75

then add all of them together and you will get the surface area
90+135+75=300
you should have an answer of 300 feet²

6 0

3 years ago

Read 2 more answers

In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study

astra-53 [7]

Answer:

C. z = 2.05

Step-by-step explanation:

We have to calculate the test statistic for a test for the diference between proportions.

The sample 1 (year 1995), of size n1=4276 has a proportion of p1=0.384.

$p_1=X_1/n_1=1642/4276=0.384$

The sample 2 (year 2010), of size n2=3908 has a proportion of p2=0.3621.

$p_2=X_2/n_2=1415/3908=0.3621$

The difference between proportions is (p1-p2)=0.0219.

$p_d=p_1-p_2=0.384-0.3621=0.0219$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{1642+1415}{4276+3908}=\dfrac{3057}{8184}=0.3735$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3735*0.6265}{4276}+\dfrac{0.3735*0.6265}{3908}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00006}=\sqrt{0.00011}=0.0107$