How do I find the area of the base of a three dimensional cone

I NEED AN ANSWER ASAP!! IM MARKING THE FIRST ANSWER BRAINLIEST!

Rom4ik [11]

The tennis balls will not go the same distance

Hope this helps :D

(Sometimes i don’t know what the question is asking so pls tell me if I’m wrong)

3 0

2 years ago

A garden hose having an internal diameter of 0.740 in. ( 1.8796 cm ) is connected to a lawn sprinkler that consists merely of an

lara31 [8.8K]

Answer:

It will leave the sprinkler at speed of $v_2=613.87m/sec$

Step-by-step explanation:

We have given internal diameter of the garden hose $d_1=0.740in=1.8796cm$

So radius $r_1=\frac{d_1}{2}=\frac{1.8796}{2}=0.9398cm$

So area $A_1=\pi r_1^2=3.14\times 0.9398^2=2.7733cm^2$

Water in the hose has a speed of 4 ft/sec

So $v_1=4ft/sec=121.92cm/sec(As\ 1ft/sec\ =30.48cm/sec)$

Number of holes n = 36

Diameter of each hole $d_2=0.1397cm$

So radius $r_2=0.0698cm$

So area $A_2=\pi r^2=3.14\times 0.0698^2=0.0153cm^2$

From continuity equation

$A_1v_1=nA_2v_2$

$2.7733\times 121.92=36\times 0.0153\times v_2$

$v_2=613.87m/sec$

7 0

3 years ago

If zach runs 6 miles, 5 days a week. How many miles does he run in 2 weeks,

Genrish500 [490]

Answer:

60 miles

Step-by-step explanation:

In 1 week Zach will run

6* 5 = 30 miles

Multiply this by 2 for 2 weeks

30*2 = 60 miles

4 0

3 years ago

How to make a mixed or whole number 7/3

eimsori [14]

Answer:

mixed number: 7/3= 2 1/3

4 0

2 years ago

Read 2 more answers

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

Let $\mu$ = average miles for deluxe tires

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, test statistics = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0

3 years ago