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kenny6666 [7]
3 years ago
8

Which of these could be the graph of F(x) = In x + 1?

Mathematics
1 answer:
mafiozo [28]3 years ago
7 0
Graph C because it goes through the point (1,0)
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<img src="https://tex.z-dn.net/?f=4x%5E%7B0%7D%20x16x%5E%7B0%7D" id="TexFormula1" title="4x^{0} x16x^{0}" alt="4x^{0} x16x^{0}"
statuscvo [17]

Answer:

64

Step-by-step explanation:

its just how it is

8 0
3 years ago
How can you show that two objects are proportional with a table?
vredina [299]
So this is our table:

               -----------------
x-values | a b c d e f  |
               -----------------
y-values | g h  i  j k l   |
               -----------------

You can know that the values in the table are proportional if:

\frac{a}{g}=\frac{b}{h} =\frac{c}{i}=\frac{d}{j}=\frac{e}{k}=\frac{f}{l}

Hope that helps :)
5 0
3 years ago
the measure of an angle is 12 less than twice the measure of its supplement. what is the measure of the angle
Alex

Answer:

Two angles are said to be supplementary if they add up to 180 degrees. So the supplement of an angle is obtained by subtracting it from 180. i.e., if x is an angle, its supplement is 180−x

7 0
4 years ago
How much will be left after 24 hrs if it’s half life is 6 hours?
Ket [755]
The quantity remaining will be 
.. 448*(1/2)^(24/6) = 448/16 = 28 . . . . grams
6 0
3 years ago
Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last
lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

H_0: \mu=1102\\\\H_a:\mu < 1102

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55

The degrees of freedom are df=599

df=n-1=600-1=599

The P-value for this test statistic is:

P-value=P(t

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is  t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0
4 years ago
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