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AlekseyPX
3 years ago
10

The probability that the bus is late on Monday is 20%, and the probability that the bus is late on Tuesday is 10%. What is the p

robability that the bus will be late on both days? Express your answer as a decimal rounded to the nearest hundredth.
Mathematics
1 answer:
Aleks [24]3 years ago
7 0
Add 20% and 10% (30%). Then, because it is two days combined, it would be 30% of 200. Divide each by 2, and there is your answer. 

<span>The probability that the bus will be late on both days is: 15%.</span>
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A shop needs to mail 77 checks to the bank if they put 4checks in each envelope how many checks will be in the final envelope
andreev551 [17]
You have to divide 77 by 4 and then the reminder will be 1 so only one check will be there in the final envelop.
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3 years ago
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Please help it due rn​
leonid [27]

Answer:

B) (1/2, -8)

Step-by-step explanation:

(1, -6) and (0, -10)

Midpoint formula:

((x1+x2)/2, (y1+y2)/2)

Solving for x:

(x1+x2)/2

(1 + 0)/2

1/2

Solving for y:

(y1+y2)/2

(-6-10)/2

(-16)/2

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8 0
2 years ago
The product of x and -8 is more than 72. What are all possible values of x?​
Fiesta28 [93]

Answer:  x < -9

Step-by-step explanation:

-8x > 72          

x < -9      when dividing a negative number in an inequality, change the sign to opposite.

the number are less than -9.

7 0
2 years ago
WHATS THE ANSWER?? answer ASAP NO FILE AT ALL
dlinn [17]

Answer:

B

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores
EastWind [94]

Answer:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

Step-by-step explanation:

We have the following info given from the problem

\bar X_1 = 52 sample mean for this year

s_1= 13 sample deviation for this year

n_1 = 75 random sample selected for this year

\bar X_2 = 49 sample mean for last year

s_2= 11 sample deviation for last year

n_1 = 53 random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df= n_1 +n_2 -2 = 75+53-2= 126

The confidence level is 0.95 and the significance would be \alpha=0.05 and \alpha/2 =0.025 so then the critical value for this case is :

t_{\alpha/2}= 1.979

The margin of error would be:

ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300

And the confidence interval would be given by:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

6 0
3 years ago
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