He spent $14.44 more on groceries this week than last week
<h3>How to determine the additional amount that was spent this week?</h3>
From the question, the given parameters are
Amount spent last week = $58.34
Amount spent this week = $72.78
The additional amount that was spent at the grocery store this week is calculated by subtracting the amount spent this week from the amount spent last week
This is represented as
Additional amount = Amount spent this week - Amount spent last week
Substitute the known values in the above equation
So, we have
Additional amount = 72.78 - 58.34
Evaluate the difference
Additional amount = 14.44
Hence, an additional of $14.44 was this week
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Answer:

Step-by-step explanation:
Hi there!
<u>What we need to know:</u>
- Linear equations are typically organized in slope-intercept form:
where <em>m</em> is the slope and <em>b</em> is the y-intercept - Parallel lines always have the same slope (<em>m</em>)
<u>Determine the slope (</u><em><u>m</u></em><u>):</u>
<u />
<u />
The slope of the given line is
, since it is in the place of <em>m</em> in y=mx+b. Because parallel lines always have the same slope, the slope of a parallel line would also be
. Plug this into y=mx+b:

<u>Determine the y-intercept (</u><em><u>b</u></em><u>):</u>

To find the y-intercept, plug in the given point (6,14) and solve for <em>b</em>:

Therefore, the y-intercept of the line is 22. Plug this back into
:

I hope this helps!
Answer:
0.0098
Step-by-step explanation:
Probability of being pregnant:100/1000
=1/10.
98% chance:=98/100 ×1/10
=98/1000
Therefore the probability that she really is pregnant is: 0.098
Answer:
The 95% confidence interval for the measurements is [48.106, 53.494].
Step-by-step explanation:
The average M of this sample is

The standard deviation s of this sample is

![s=\sqrt{\dfrac{1}{4}\cdot [(52-50.8)^2+(48-50.8)^2+(49-50.8)^2+(52-50.8)^2+(53-50.8)^2]}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B4%7D%5Ccdot%20%5B%2852-50.8%29%5E2%2B%2848-50.8%29%5E2%2B%2849-50.8%29%5E2%2B%2852-50.8%29%5E2%2B%2853-50.8%29%5E2%5D%7D)
![s=\sqrt{\dfrac{1}{4}\cdot [(1.44)+(7.84)+(3.24)+(1.44)+(4.84)]}=\sqrt{\dfrac{18.8}{4}}=\sqrt{4.7}\\\\\\s=2.168](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B4%7D%5Ccdot%20%5B%281.44%29%2B%287.84%29%2B%283.24%29%2B%281.44%29%2B%284.84%29%5D%7D%3D%5Csqrt%7B%5Cdfrac%7B18.8%7D%7B4%7D%7D%3D%5Csqrt%7B4.7%7D%5C%5C%5C%5C%5C%5Cs%3D2.168)
The degrees of freedom are

Then, the critical value of t for a 95% CI and 4 degrees of freedom is t=2.776.
The margin of error of the CI is:

Then, the lower and upper bounds of the CI are:

The 95% confidence interval for the measurements is [48.106, 53.494].