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3 years ago

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On a street, each house has an address between 1000 and 1099, inclusive. at least how many houses are there if at least two of t

hem have addresses that are consecutive integers

1 answer:

notsponge [240]3 years ago

8 0

There are at least 3 houses on a street if at least two of them have addresses that are consecutive integers.

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If $1000 is invested at 2% interest, compounded continuously, for 5 years what is the ending balance? Round to the nearest cent

Radda [10]

1104.08 is your answer for this question

7 0

3 years ago

Simplity (4^-2)^5<br> A.1/4^10<br> B. 4^3<br> C.1/4^3<br> D. 4^10

Goshia [24]

Answer:

$(4^{-2})^5=\dfrac{1}{4^{10}}$

Step-by-step explanation:

The given expression is : $(4^{-2})^5$

We need to simplify the above expression.

We know that, $x^{-a}=\dfrac{1}{x^a}$

or

$(4^{-2})^5=(\dfrac{1}{4^2})^5\\\\=\dfrac{1^5}{(4^2)^5}\\\\\because (x^b)^c=x^{b\times c}\\\\=\dfrac{1}{4^{10}}$

So, the simplified form of the given expression is $\dfrac{1}{4^{10}}$ . Hence, the correct option is (A).

5 0

3 years ago

Read 2 more answers

A ditch contains 10 centimeters of water. Rainwater accumulates in the ditch as follows: 10 centimeters of water by the end of t

kondaur [170]

Firstly, let's create a function of f(t) where t represents the time that has past, and f(t) represents the amount of rainwater. We know that when t=1, then f(t)=10, and t=2 then f(t)=15. So, let's take that and analyze it:

(1,10)
(2,15)
m = (15-10)/(2-1) = 5
y-intercept = 5
∴ f(t) = 5t+5

Now we just evaluate t for 10:

f(10) = (5*10)+5
f(10) = 55

3 0

3 years ago

Based on historical data, your manager believes that 40% of the company's orders come from first-time customers. A random sample

Vedmedyk [2.9K]

Answer:

$P(0.26 \leq p \leq 0.43)=0.7204-0.0032=0.7172$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The population proportion have the following distribution

$p \sim N(p=0.4,\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4(1-0.4)}{91}}=0.0514)$

And we can solve the problem using the z score on this case given by:

$z=\frac{p_o -p}{\sqrt{\frac{p(1-p)}{n}}}$

We are interested on this probability:

$P(0.26 \leq p \leq 0.43)$

And we can use the z score formula, and we got this:

$P(\frac{0.26 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}} \leq Z \leq \frac{0.43 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}})$

$P(-2.726 \leq Z \leq 0.584)$

And we can find this probability like this:

$P(-2.726 \leq Z \leq 0.584)=P(Z$

7 0

3 years ago

John’s friend told him that he could earn $49 for handing out flyers at a local concert John wants to calculate the hourly rate

son4ous [18]

$\frac{3.5x}{3.5} = \frac{49}{3.5}$
$x = 14$

5 0

4 years ago