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2 years ago

11

Finn and Ellie sell oranges at a produce stand. Finn earns $5 for each crate of oranges he sells. At the end of the week, Ellie

has earned $20 more than Finn. The following expression shows Ellie's earnings:
20 + 5z

In the expression, what does the second term represent?

1 answer:

Alex777 [14]2 years ago

6 0

Answer:

4

Step-by-step explanation:

4x5=20

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You have a need to purchase 30 units of a product and have them delivered to your company in five days. The sale price is $35 ea

tino4ka555 [31]

Answer: $1411.50

Step-by-step explanation:

Since the sale price is $35 each and there are 30 units, the cost will be:

= $35 × 30

= $1050

We then add the sales tax on the product which is 8%. This will be:

= $1050 + (8% × $1050)

= $1050 + (0.08 × $1050)

= $1050 + $84

= $1134

We then add Shipping price which is $15. This will be:

= $1134 + $15

= $1149

We then add the 25% rush charge on the sales price. This will be:

= $1050 × 25%

= $1050 × 0.25

= $262.50

To get total cost, this will be:

= $1149 + $262.50

= $1411.50

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2 years ago

Read 2 more answers

Two well-known aviation training schools are being compared using random samples of their graduates. It is found that 70 of 140

egoroff_w [7]

Answer:

Step-by-step explanation:

Given that two well-known aviation training schools are being compared using random samples of their graduates

Fly more academy 70 of 140

Blue Yonder 104 of 260

Combined pass = (70+104) out of (140+260)

a) Pooled proportion= $\frac{174}{400} \\=0.435$

b) H0: p1 = p2

Ha: p1 ≠p2

(two tailed test)

p difference= $\frac{70}{140} -\frac{104}{260} =0.10$

std error for difference (using pooled proportion) = $\sqrt{\frac{0.435*0.565}{400} } \\=0.0248$

Test statistic = p difff/std error = 4.034

c) Critical value for 0.05 is 1.96

d) p value is < 0.005

Since p < 0.05 our significant level we reject H0

There is significant difference between the two proportions.

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2 years ago

For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f

mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters

Step-by-step explanation:

<u>Step 1</u>:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution $P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}$

<u>Step 2:</u>-

The probability that no flaws occur in a certain portion of wire

$P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}$

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

<u>Conclusion</u>:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

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2 years ago

If someone wants to ask a question that doesn't have a lot is variability, which answer should it be?

nikklg [1K]

I believe its C or D

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2 years ago

For circle O, and m∠ABC = 55°. In the figure, ∠_____ and ∠____ have measures equal to 35°.

luda_lava [24]

Answer:

In the figure ∠ABO and ∠BCO have measures equal to 35°.

Step-by-step explanation:

<u><em>The complete question is</em></u>

For circle O, m CD=125° and m∠ABC = 55°

In the figure<____, (AOB, ABO, BOA) and <_____ (BCO, OBC,BOC) have measures equal to 35°

The picture in the attached figure

step 1

Find the measure of angle COB

we know that

$m\angle COB=arc\ CD$ ----> by central angle

we have

$arc\ CD=125^o$

therefore

$m\angle COB=125^o$

step 2

we know that

AB is a tangent to the circle O at point A

so

ABC and ABO are right triangles

In the right triangle ABC

Find the measure of angle BCA

Remember that

$m\angle BCA+m\angle\ ABC=90^o$ ---> by complementary angles in a right triangle

we have

$m\angle ABC=55^o$

substitute

$m\angle BCA+55^o=90^o$

$m\angle BCA=90^o-55^o=35^o\\$

step 3

In the triangle BCO

Find the measure of angle CBO

we know that

$m\angle CBO+m\angle COB+m\angle BCO=180^o$ ---> the sum of the interior angles in any triangle must be equal to 180 degrees

we have

$m\angle COB=125^o$

$m\angle BCO=m\angle BCA=35^o$ -----> have measure equal to 35 degrees

substitute

$m\angle CBO+125^o+35^o=180^o$

$m\angle CBO=180^o-160^o=20^o$

step 4

Find the measure of angle ABO

In the right triangle ABO

we know that

$m\angle ABC=m\angle CBO+m\angle ABO$ ----> by angle addition postulate

we have

$m\angle ABC=55^o$

$m\angle CBO=20^o$

substitute

$55^o=20^o+m\angle ABO$

$m\angle ABO=55^o-20^o=35^o$ ----> have measure equal to 35 degrees

therefore

In the figure ∠ABO and ∠BCO have measures equal to 35°.

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2 years ago