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Nina [5.8K]
4 years ago
15

If Patricia drives at two-third of her usual speed, she covers a certain distance in one hour more than the time she takes while

driving at her usual speed. The time taken by her to cover this distance with her usual speed is _____ hours.
please help me!
Mathematics
1 answer:
zalisa [80]4 years ago
3 0
Let r = usual driving rate
let t = usual driving time
We need to figure out t

The distance she covers in her usual time at her usual rate is r*t

The distance she covers in her new time at her new rate is:
(1+t)*((2/3)r)

Set this equal to each other and solve for t.

rt = (2/3)r + (2/3)rt
(1/3)rt = (2/3)r
(1/3)t = (2/3)
t = 2

So her usual time is 2 hours. (There's probably a faster way to do this)
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The marketing director of a large department store wants to estimate the average number of customers who enter the store every f
Dahasolnce [82]

Answer:

36.5674\leq x'\leq61.4326

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, \alpha is equal to 0.05 (it is calculated as: 1 - 0.95) and  t_{\alpha /2} is the t value with n-1 degrees of freedom that let a probability of \alpha/2 on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and t_{\alpha /2} by 2.2621 we get:

49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326

Finally, the interval values that she get is:

36.5674\leq x'\leq61.4326

8 0
3 years ago
A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is \frac{1}{13}.

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is 4\times 51!.

The probability that the last card dealt is an ace is

P=\frac{4\times 51!}{52!}

P=\frac{4\times 51!}{52\times 51!}

P=\frac{4}{52}

P=\frac{1}{13}

Therefore, the probability that the last card dealt is an ace is \frac{1}{13}.

4 0
3 years ago
A sea lion dove from the water's surface at sea level to an altitude of −216.12 meters in 2.4 minutes. What is the average chang
Rudik [331]
Reduce the numbers by dividing both by 2.4 to get the distance traveled per min which is 90.05 meters a min, so every min its altitude decreases by 90.05 meters
6 0
3 years ago
Ling must spend no more than $40.00 on decorations for a party. She has spent $10.00 on streamers and wants to buy bags of ballo
andreyandreev [35.5K]

Answer:

She can buy from 0 to 20 bags, but no more.

8 0
3 years ago
Anybody can help me?
irinina [24]

Graph C depicts the relation between No. of sweaters ( Nombre de chandails vendus ) and the fund raised in the campaign (Campagne de financements )

<h3>What is a Graph ?</h3>

Graph is the statistical representation of data , It helps to observe large data systematically.

It is given that

Sale of one sweater gives $3

The relation between the amount raised for funding , M and the number of shirts sold , n is given by

M = 3n

It has to be found that which graph exactly depicts the relation and the condition as given

Graph C depicts the relation given as , The graph is between No. of sweaters ( Nombre de chandails vendus ) and the fund raised in the campaign (Campagne de financements ) , and for every 100 sweater , $300 is earned .

Therefore Graph C is the answer.

To know more about Graph

brainly.com/question/14375099

#SPJ1

6 0
2 years ago
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