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Andrew [12]
3 years ago
8

I need help on these questions 1-5 please help

Mathematics
1 answer:
Evgen [1.6K]3 years ago
7 0
The answer for number one is d because 7 multiply by5 is 35
The answer for number two is a
You might be interested in
Factor each completely. <br><br>10a² - 9a + 2 <br><br>​
Sophie [7]

Answer:

a= \frac{2}{5} or \frac{1}{2}

Step-by-step explanation:

Equate the equation to zero

10a^{2}-9a+2=0

10a^{2}-5a-4a+2=0

5a(2a-1)-2(2a-1)=0

Write out the factors

(5a-2)(2a-1)=0

Then equate each factor to zero

5a-2=0 OR 2a-1=0

5a=2  OR  2a=1

a= \frac{2}{5}   OR a= \frac{1}{2}

a=\frac{2}{5} or \frac{1}{2}

3 0
3 years ago
A noncontinuous loop that prevents the flow of current is a(n)<br> circuit.
Anastaziya [24]

Answer:

It is an open circuit, yes

Step-by-step explanation:

5 0
4 years ago
Alex borrowed $12.50 from his friend Danilo. He paid him back $8.75. How much does he still owe? *
ollegr [7]

Answer:

$3.75

Step-by-step explanation: $12.50

                                             $  8.75

                                           ________

                                              $  3.75

8 0
3 years ago
Choose the inequality that represents the following graph.
DaniilM [7]
The answer is b your welcome
3 0
3 years ago
Read 2 more answers
Assume that the number of customers who arrive at a water ice stand follows the Poisson distribution with an average rate of 6.4
Nady [450]

Answer:

18.88% probability that three or four customers will arrive during the next 30 minutes

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

Average rate of 6.4 per 30 minutes.

This means that \mu = 6.4

What is the probability that three or four customers will arrive during the next 30 minutes?

P = P(X = 3) + P(X = 4)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 3) = \frac{e^{-6.4}*(6.4)^{3}}{(3)!} = 0.0726

P(X = 4) = \frac{e^{-6.4}*(6.4)^{4}}{(4)!} = 0.1162

P = P(X = 3) + P(X = 4) = 0.0726 + 0.1162 = 0.1888

18.88% probability that three or four customers will arrive during the next 30 minutes

4 0
3 years ago
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