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2 years ago

Please answer correctly, will give brainlist.

Mathematics

1 answer:

aliina [53]2 years ago

8 0

Answer:

the answer is 77

Step-by-step explanation:

A measures 32 and B measures 109 so you subtract 109 and 32 and your answer will be 77

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A service center receives an average of 0.6 customer complaints per hour. Management's goal is to receive fewer than three compl

True [87]

Answer:

0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.

Step-by-step explanation:

We have the mean during a time-period, which means that the Poisson distribution is used to solve this question.

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 interval.

A service center receives an average of 0.6 customer complaints per hour.

This means that $\mu = 0.6h$ , in which h is the number of hours.

Determine the probability that exactly four complaints will be received during the next eight hours.

8 hours means that $h = 8, \mu = 0.6(8) = 4.8$ .

The probability is P(X = 4).

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

$P(X = 4) = \frac{e^{-4.8}*4.8^{4}}{(4)!} = 0.18203$

0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.

5 0

2 years ago

John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created

Over [174]

We will have that for the previous year, he had in the account:

$m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106$

For the year previous to that one, he had:

$m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964$

For the previous year to that, he had:

$m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462$

For the year previous to that one, he had:

$m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314$

For the second year that the money was in the account, there was:

$m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775$

And the orinial ammount of money that was in the account was:

$m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669$

From this, we know that Jhon had originally $1400.37 in the bank account.

6 0

2 months ago

Indicate whether the study is an observational study or a controlled experiment. A group of boys is randomly divided into two gr

MariettaO [177]

Answer:

Controlled Experiment

Step-by-step explanation:

The controlled experiment involves the experiment in which all factors are considered as constants whereas one factor is not fixed and that one is independent variable whose effect is to be tested. In short, in controlled experiment one group of data contains control group and one or more other groups are examined by considering all other factors constant except for one factor. The control group is the group that doesn't receive any treatment whereas treatment is applied to experimental group and then effect of treatment is measured through comparing both groups.

Here group of boys is divided in to two groups

The group watched violent cartoon= Experimental group

The group watched cartoons without violence= Control group

The researchers want to know whether the violent actions of boys are linked to violent cartoon and for this purpose these boys are divided into experimental group and control group and after 1 hour of cartoons watching the researcher will observed boys for next two hours.

Hence the study is Controlled experiment.

6 0

2 years ago

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z

Roman55 [17]

Answer:

$(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}$

Step-by-step explanation:

Given system of equations:

$\begin{cases}y=x^2-9\\y=4x-4\end{cases}$

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

$\begin{aligned}x^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}$