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3 years ago

12

3x + 2y = 2

2 answers:

Ivanshal [37]3 years ago

8 0

I feel like the answer is A. (-2,4) :o cause I tried to get D. which didn't work so I solved it in a weird way that you wouldn't understand cause i don't want to confuse you at all. So i'm leaning on the A. :/

if this is wrong please don't hate me >< I tried 3 times

~ShadowLePanda

HACTEHA [7]3 years ago

5 0

I hope this helps you

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A triangle has sides with lengths of 5 millimeters, 12 millimeters, and 14 millimeters. Is it a right triangle?

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3 years ago

A random sample of 11 fields of rye has a mean yield of 20.1 bushels per acre and standard deviation of 7.66 bushels per acre. D

Lady bird [3.3K]

Answer:

17.1≤x≤23.1

Step-by-step explanation:

The formula for calculating the confidence interval is expressed as;

CI = x ± z*s/√n

x is the mean yield = 20.1

z is the 80% z-score = 1.282

s is the standard deviation = 7.66

n is the sample size = 11

Substitute

CI = 20.1 ± 1.282*7.66/√11

CI = 20.1 ± 1.282*7.66/3.3166

CI = 20.1 ± 1.282*2.3095

CI = 20.1 ±2.9609

CI = (20.1-2.9609, 20.1+2.9609)

CI = (17.139, 23.0609)

hence the required confidence interval to 1dp is 17.1≤x≤23.1

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3 years ago

6. Ed drives from Jefferson to Holden, a distance of 250 miles. He then travels on to

andreev551 [17]

Answer:

The constant speed with Ed is travelling is 60 mile per hour

Step-by-step explanation:

Given as :

The distance cover to travel from Jefferson to Holden is (D1) 250 miles

The distance cover to travel from Holden to Paxton is (D2) 50 miles

Total time taken to complete entire journey = T = 5 hours

Let The constant speed of Ed = S mph

Time = $\frac{Distance}{Speed}$

Or, Time = $\frac{D1}{S}$ + $\frac{D2}{S}$

Or, 5 hour = $\frac{250}{S}$ + $\frac{50}{S}$

Or, 5 × S = 250 + 50

So, 5 × S = 300

∴ S = $\frac{300}{5}$

I.e S = 60 mph

Hence , The constant speed with Ed is travelling is 60 mile per hour Answer

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Step-by-step explanation:

Because both seem to have outliers, we want to use IQR instead of standard deviation to determine the spread of data.

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