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olya-2409 [2.1K]
2 years ago
12

(15) ? Is this correct or am I missing something, thanks :)

Mathematics
1 answer:
svetoff [14.1K]2 years ago
4 0
You are correct!

Step 1:Write what you know
Madeline's Height: 6 ftMadeline's Shadow: 108 in
Friend's Height: X (we use X as a variable because his height is unknown)Friend's Shadow: 108 in - 1.5 ft                              108 in - 18 in                                 90 in
Step 2:Set up proportions
6          x----- = -----108      90 
Step 3:Cross Multiply & Solve
108x = 540x = 5 ft tall

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5 0
2 years ago
Frrrrrrrrreee points​
salantis [7]
Thanks! :)

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larisa86 [58]

Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form:

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Decimal Form:

c=0.25

8 0
3 years ago
Lily bought 25.29 pounds of grapefruit. The lightest grapefruit weighed 1.4 pounds. The heaviest grapefruit weighed 1.6 pounds.
Gemiola [76]

Answer:

B

Step-by-step explanation:

GIven that the lightest is 1.4 and the heaviest is 1.6, this implies that ALL the fruit must weigh between 1.4 and 1.6.

This also implies that the average weight of the fruit must also be between 1.4 lb and 1.6 lb.

recall that : average weight = total weight / number of fruit

we are given than total weight = 25.29 lb, so

average weight = 25.29 / number of fruit

Just go down the choices and test which choice gives an average weight between 1.4 and 1.6 lbs.

Choice A: number of fruit = 19

average = 25.29 / 19 = 1.33  (outside of range, not the answer)

Choice B: number of fruit = 17

average = 25.29 / 17 =1.49   (within range , possible answer)

Choice C: number of fruit = 10

average = 25.29 / 10 = 2.529 (outside of range, not the answer)

Choice D: number of fruit = 50

average = 25.29 / 50 = 0.5058  (outside of range, not the answer)

From the above, we can see that only B gives an average between 1.4 and 1.6 lb.

5 0
3 years ago
Statistics question; please help.
Alex_Xolod [135]

Answer:

Does this sample provide convincing evidence that the machine is working properly?

Yes.

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$z=\frac{x- \mu }{ \frac{\sigma}{\sqrt{n}} }=\frac{ (x-\mu)\sqrt{n}}{\sigma}  $

Where,

x: \text{ sample mean}

\sigma: \text{ standard deviation}

n: \text{ sample size determination}

\mu: \text{ hypothesized size of the screw}

$z=\frac{(1.476-1.5)\sqrt{200} }{0.203 } $

$z=\frac{(-0.024)10\sqrt{2} }{0.203 } $

z \approx -1.672

Once the significance level was not given, It is usually taken an assumption of a 5% significance level.

Taking the significance level of 5%, which means a confidence level of 95%, we have a z-value of \pm 1.96

Therefore, we <u>fail to reject the null</u>. It means that the hypothesis test is not statistically significant: the average length is not different from 1.5!

7 0
2 years ago
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