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nlexa [21]
3 years ago
5

Alex wrote checks on Tuesday for $35 and $40. He also made a deposit in his checking account of $170. Find the overall change in

the amount in his checking account.
Mathematics
2 answers:
Eva8 [605]3 years ago
3 0

Answer: He retrieved 75$ from his account then he put back 170$ so +75 and -170 so in total the answer would be 245$

kow [346]3 years ago
3 0

Answer:  He would have $100 in his bank account from the information provided .

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Disregarding wind resistance the distance a body falls from rest varies directly as th squares of the time it falls if a skydive
Firdavs [7]

Answer:

The skydiver will fall 1600 ft in 10 seconds.

Step-by-step explanation:

<u>Distance a body falls from rest varies directly as the squares of the time it falls </u>

Let distance = s    

      time = t

Therefore , s = k \times t^{2}

Given, the body falls 64 ft in 2 seconds.

Substituting in above equation,

64 = k \times (2)^{2}

64 = k \times (4)

 k  = \frac{64}{4}

 k = 16 ft/s^{2}

We have to find how much he will fall in 10 seconds.

s = k \times t^{2}

s = 16 \times 10^{2}

s = 16 \times 100

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3 years ago
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iren2701 [21]
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3 years ago
A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21
sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{340119}{10} = 34011.9

Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821

S.D = \sqrt{\frac{2711418821}{9}} = 17357.09

Confidence interval:

\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621

34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)

7 0
3 years ago
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Answer:

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Hope this helps!

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