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2 years ago

5

Suppose y varies directly with x. Write a direct variation equation that relates x and y.

1 answer:

timama [110]2 years ago

8 0

$\huge\mathfrak\purple{Answer:-}$

★ <u>Option (A) y = -5x</u> is the right answer.

<h2><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h2>

<h3>I hope this helps! :)</h3>

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An airplane is preparing to land at an airport. It is 53100 feet above the ground and is descending at the rate of 3300 feet per

Crank

9514 1404 393
Answer:
after 7 minutes
19,600 feet
Step-by-step explanation:
Here's the "pencil and paper" solution:
The two altitude equations are ...
y = 41300 -3100x
y = 2800x
They can be solved by setting the expressions for y equal to each other.
2800x = 41300 -3100x
5900x = 41300
x = 41300/5900 = 7
y = 2800·7 = 19600
The planes will both be at 19,600 feet after 7 minutes.

7 0

2 years ago

A wood stove burns 4 logs in 2 hours how many logs burn in 9 hours ?

ziro4ka [17]

Answer:

18 logs.

Step-by-step explanation:

Since a stove burns 4 logs per 2 hours, then the stove will burn 2 logs in one hour. There fore in 9 hours the stove will burn 2*9 which is 18 logs.

4:2= 2 logs per hour

2*9=18 logs per 9 hours.

8 0

2 years ago

Read 2 more answers

Please help and please give explanation

Yuki888 [10]

Answer:

cylinder surface area = 201.06in²

surface area of cone = 56.56in.²

Step-by-step explanation:

cylinder surface area formula = 2πrh + 2πr²

refer the pic

cone surface area formula = πr ( r + √h² + r² )

refer the sec pic

surface area of the solid

= ( 201.06 + 56.56 ) in.²

= 257.62in.²

= 258in.²

4 0

1 year ago

Answer number 3<br><br><br> quickly i am in a rush

SVEN [57.7K]

Answ34.5

Step-by-step explanation:

8 0

2 years ago

A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty

Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: $\hat{p}=0.84$

Significance level : $\alpha= 1-0.98=0.02$

Critical z-value : $z_{\alpha/2}=2.33$ (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}$

i.e. $\approx0.84\pm 0.085$

i.e. $(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)$

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0

2 years ago