All categories

3 years ago

A go-kart's speed is 607,200 ft per hour. What is the speed in miles per hour?

Mathematics

1 answer:

Inessa [10]3 years ago

4 0

There are 5280 feet in a mile. So divide 607,200 feet by 5280 to get 115. The go-kart's speed is 115 miles per hour.

You might be interested in

Maggie takes three electives: art, Spanish, and creative writing. She calculated the mean absolute deviation of the points she e

stiv31 [10]

Answer:

i think its art

Step-by-step explanation:

6 0

2 years ago

7 - 12c = 43 show work and check plz

Svetradugi [14.3K]

The answer is -3

-12c = 43 - 7
-12c = 36
÷ -12 ÷-12
c = -3

4 0

2 years ago

Exercise questions

balu736 [363]

Answer:

a)1

b)1

c)1

Step-by-step explanation:

There's some identity trigonometric equation, which are valid for all angles,and they doesn't depends on the measure of angle!

some of em are follows:. (x is the given angle)

sin(x)^2+cos(x)^2=1
cosec(x)^2=1+cot(x)^2
sec(x)^2=1+tan(x)^2

You can remember these identity, its gonna help alot.

now back to question,. {x is representing angles)

for (a) sin(x)^2+cos(x)^2=1, this is true for all x, dat means that for all the angle given in question(for ,15°,30°,45°,60° and 120°),we will get 1

for(b) ,

cosec(x)^2=1+cot(x)^2

i.e, cosec(x)^2-cot(x)^2=1, again this is true for all x dat means that for all the angle given in question ,we will get 1

for (c),

sec(x)^2=1+tan(x)^2

i.e,sec(x)^2-tan(x)^2=1,again this is true for all x, dat means that for all the angle given in question ,we will get 1

✌️:)

3 0

2 years ago

A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies

RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean weight = 733 grams

$\sigma$ = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

P(X > x) = 0.05 {where x is the required weight}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-733}{9}$ ) = 0.05

P(Z > $\frac{x-733}{9}$ ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

$\frac{x-733}{9}=1.645$

${x-733}}=1.645\times 9$

$x = 733 + 14.81$

x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0

2 years ago

(repost cuz the last didn’t load) please help ‼️

shtirl [24]

1: (pi)5^2 = 78.54 then multiply that by 11 and you get 863.94
2: (pi)13^2= 530.93 then multiply that by 8 and you get 47247.44
3. (pi)18^2= 1017.88 then multiply that by 22 and you get 22393.36

8 0

3 years ago