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Sloan [31]
2 years ago
6

This we havent learned but teacher gave us for end of year est plz help

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
3 0

Answer:

15

Step-by-step explanation:

Finding total number of pupils(y)

3/y x 360°=27°

3x360°/y=27°

1080°/y=27°

1080°=27°y

1080°/27°=27°y/27°

40=y

Number of pupils who said that Maths was their favorite (z)

z/40x360°=135°

360°z/40=135°

360°z=135°x40

360°z=5400°

360°z/360°=5400°/360°

z=15

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(a) 23,45,678

(b) 56,78,090

<em><u>2</u></em><em><u>.</u></em> <em><u>International</u></em><em><u> </u></em><em><u>System</u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

(a) 234,589

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8 0
3 years ago
This Line Chart shows the number of soft drink bottles a vendor sold during each month of baseball season.
navik [9.2K]
The answer is A: 50



400 - 350 = 50
4 0
2 years ago
Which rule describes the transformation? picture below​
Mekhanik [1.2K]

Answer:

Try D) rotation 180 degrees about the origin.

Step-by-step explanation:

When you look at it, it appears to move 180 degrees about that origin.

Hope this helps! Let me know!

6 0
3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
A jet’s speed in still air is 240 mph. One day it flew 700 miles with a tailwind (the wind pushing it along) and then returned t
Bess [88]

Speed of the wind for jet speed in still air 240mph and jet covers 700 miles with tailwind and same distance against the wind in total time of 6 hours is equal to 40mph.

As given in the question,

Given data:

Jet speed in still air = 240mph

Let x be the speed of the wind.

Speed with tail wind = 240 +x

Distance covered with tail wind = 700miles

Time taken by jet with tail wind

= Distance/ speed

= 700 / (240 +x)  __(1)

Speed against the wind = 240 - x

Distance covered against the wind = 700miles

Time taken by jet against the wind

= Distance/ speed

= 700 / (240 - x)  __(2)

Total time taken is 6 hours

[700 / (240 +x)] + [700 / (240 - x)] = 6

⇒ 700( 240 -x + 240 + x) = 6 (240 +x)(240 - x)

⇒ 700 ( 480) = 6 ( 240² -x²)

⇒ 700 (80) = 57600 -x²

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⇒ x² = 1600

⇒ x = √1600

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Therefore, speed of the wind for jet speed in still air 240mph and jet covers 700 miles with tailwind and same distance against the wind in total time of 6 hours is equal to 40mph.

Learn more about speed here

brainly.com/question/28224010

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6 0
1 year ago
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