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4 years ago

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For a scavenger hunt, Jim's mom distributed a bag of 460 jelly beans evenly into 20 plastic containers and hid them around the y

ard. If, after the hunt, Jim has a total of 207 jelly beans, then how many of the plastic containers did he find?

1 answer:

Veseljchak [2.6K]4 years ago

7 0

I think he found nine bags

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The number of potholes in any given 1-mile stretch of freeway pavement in Pennsylvania has a Normal distribution. This distribut

Iteru [2.4K]

Answer:

83.85% of 1-mile long roadways with potholes numbering between 22 and 58

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 49

Standard deviation = 9

Using the Empirical Rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 22 and 58?

22 = 49 - 3*9

So 22 is three standard deviations below the mean.

Since the normal distribution is symmetric, 50% of the measures are below the mean and 50% are above the mean.

Of those 50% which are below the mean, 99.7% of those are within 3 standard deviations of the mean, that is, greater than 22.

58 = 49 + 9

So 58 is one standard deviation of the mean.

Of those which are above the mean, 68% are within 1 standard deviation of the mean, that is, lesser than 58.

Then

0.997*0.5 + 0.68*0.5 = 0.8385 = 83.85%

83.85% of 1-mile long roadways with potholes numbering between 22 and 58

5 0

3 years ago

Evaluate this pls quiick i have 18 min left online exams help<br><br> -0.0004 x 0.02

lions [1.4K]

Given:

The expression is:

$0.0004\times 0.02$

To find:

The value of the given expression.

Solution:

We have,

$0.0004\times 0.02$

It can be written as:

$=\dfrac{4}{10000}\times \dfrac{2}{100}$

$=\dfrac{8}{1000000}$

$=0.000008$

Therefore, the value of the given expression is 0.000008.

3 0

3 years ago

VXYZ is a kite. What is the length of XY?

olga nikolaevna [1]

Answer:

$5x - 6 = 3x + 4 \\ 5x - 3x = 4 + 6 \\ 2x = 10 \\ x = 10 \div2 \\ x = 5 \\ length \: of \: xy = 3 \times 5 + 4 \\ = 15 + 4 \\ = 19$

8 0

3 years ago

Read 2 more answers

Find dy/dx by implicit differentiation.

kow [346]

dy/dx by implicit differentiation is cos(πx)/sin(πy)

<h3>How to find dy/dx by implicit differentiation?</h3>

Since we have the equation

(sin(πx) + cos(πy)⁸ = 17, to find dy/dx, we differentiate implicitly.

So, [(sin(πx) + cos(πy)⁸ = 17]

d[(sin(πx) + cos(πy)⁸]/dx = d17/dx

d[(sin(πx) + cos(πy)⁸]/dx = 0

Let sin(πx) + cos(πy) = u

So, du⁸/dx = 0

du⁸/du × du/dx = 0

Since,

du⁸/du = 8u⁷ and

du/dx = d[sin(πx) + cos(πy)]/dx

= dsin(πx)/dx + dcos(πy)/dx

= dsin(πx)/dx + (dcos(πy)/dy × dy/dx)

= πcos(πx) - πsin(πy) × dy/dx

So, du⁸/dx = 0

du⁸/du × du/dx = 0

8u⁷ × [ πcos(πx) - πsin(πy) × dy/dx] = 0

8[(sin(πx) + cos(πy)]⁷ × (πcos(πx) - πsin(πy) × dy/dx) = 0

Since 8[(sin(πx) + cos(πy)]⁷ ≠ 0

(πcos(πx) - πsin(πy) × dy/dx) = 0

πcos(πx) = πsin(πy) × dy/dx

dy/dx = πcos(πx)/πsin(πy)

dy/dx = cos(πx)/sin(πy)

So, dy/dx by implicit differentiation is cos(πx)/sin(πy)

Learn more about implicit differentiation here:

brainly.com/question/25081524

#SPJ1

6 0

2 years ago

Omar rented a truck for one day. There was a base fee of $17.95, and there was an additional charge of 98 cents for each mile dr

stiv31 [10]

The given question is incomplete. The complete question is:

Omar rented a truck for one day. There was a base fee of $17.95, and there was an additional charge of 98 cents for each mile driven. Omar had to pay $23 when he returned the truck. For how many mile did he drive the truck?

Answer: Omar drove the truck for 5.15 miles

Step-by-step explanation:

Base fee = 17.95 $

Additional charge per mile = 98 cents = 0.98 $ ( 100cents = 1$)

Now Omar payed = 22 $

Let the miles he travelled = x

Now , $17.95+0.98\times x=23$

Solvimg for x :

$x=5.15miles$

Thus Omar drove the truck for 5.15 miles

5 0

3 years ago