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3 years ago

5

The length of a social media interaction is normally distributed with a mean of three minutes and a standard deviation of 0.4 mi

nutes. what is the probability that an interaction last longer than four minutes

1 answer:

alexandr1967 [171]3 years ago

3 0

It depends on Y and x cause school said so

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Please help please help please help

uranmaximum [27]

It should be 986m

4(17x6)= 408
2(17x17)= 578

408+578=986

7 0

3 years ago

Anyone wana be my friend if you do lets zoom and get stuff doneh

GalinKa [24]

Answer:

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Step-by-step explanation:

.

7 0

2 years ago

Read 2 more answers

Here is a list of ages (years) of children in a room <br><br> 8,7,5,4,7,7,2<br><br> State the median

Lina20 [59]

Answer:7

Step-by-step explanation:

To find the median you need to first put the number in order from least to greatest this would be 2, 4 ,5, 7, 7, 7, 8

next you have to find the number in the middle

That is your median, 7 is your median.

6 0

2 years ago

What is the sum of the measures of the exterior angles in a nonagon? explain

DedPeter [7]

The sum of the measures of all exterior angle in a polygon with n sides is always $360^{\circ}$ .

Nonagon is a polygon with 9 sides. If this nonagon is regular or equiangular, then all interior angle are congruent, and therefore all exterior angles are congruent. Then the measure of each exterior angle is $\dfrac{360^{\circ}}{9} =40^{\circ}$ .

5 0

2 years ago

A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro

pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

3 0

3 years ago