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Montano1993 [528]

3 years ago

7

Jim works overtime to save $500 per month for a pickup truck. The savings account Jim uses earns him 5% interest, compounded mon

thly. In two years, Jim will have in savings. He will have earned in interest on his savings.

2 answers:

Rama09 [41]3 years ago

8 0

Answer - the answer is 593 !

valentina_108 [34]3 years ago

4 0

Answer:

$1615

Step-by-step explanation:

500(1+.05/1)^24

500(1+.05)^24

500*1.05^24

500*3.23

1615

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kotykmax [81]

Answer:

Given points are ( 10 ,34 ) and ( -1 , 10)

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<em> </em><em>=</em><em>2</em><em>4</em><em>/</em><em>1</em><em>1</em>

<em>Now </em><em>the </em><em>equation </em><em>is</em>

<em>y </em><em>-</em><em> </em><em>y1 </em><em>=</em><em> </em><em>m </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>x1) </em>

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3 0

3 years ago

Keisha swims 18 ft from one side of her circular pool to the other, swimming straight through the center. She then swims one tim

elena-s [515]

Answer:

Swimming one time around the pool is 56.5 ft

Step-by-step explanation:

Swimming 18 ft from one side of her circular pool to the other, swimming straight through the center means the circular swimming pool has a diameter of d=18 ft.

The circumference of the circle is calculated using:

$c = \pi \: d$

where d=18 is the diameter and π=3.14 as given.

We substitute the values to get:

$c = 3.14 \times 18$

$c = 56.52$

We approximate to the nearest tenth to get:

$c = 56.5ft$

6 0

3 years ago

In preparing a certain ointment, a pharmacist used 28.35 g of zinc oxide instead of the 31.1 g called for. Calculate the percent

Vilka [71]

Answer: 9.70%

Step-by-step explanation:

The formula to find the percentage error is given by :-

$\%\text{ error}=\dfrac{|\text{Estimate-Actual}|}{\text{Actual}}\times100$

Given : Actual mass of zinc oxide used by pharmacist = 28.35 g

Estimated mass of zinc oxide used by pharmacist =31.1 g

Now, $\%\text{ error}=\dfrac{|31.1-28.35|}{28.35}\times100$

i.e. $\%\text{ error}=\dfrac{2.75}{28.35}\times100$

i.e. $\%\text{ error}=9.70017636684\approx9.70\%$

Hence, the percentage of error on the basis of the desired quantity.= 9.70%

5 0

3 years ago

A group of students were given a spelling test the table shows their mark Mark: 6,7,8,9,10 frequency:5,4,7,10,4 a) work out the

____ [38]

Answer:

Step-by-step explanation:

From the information given,

Mark: 6,7,8,9,10

frequency:5,4,7,10,4

a) Range = highest mark - lowest mark

Range = 10 - 6 = 4

b) The number of students in the group is the sum of the frequency. Therefore,

Number of students = 5 + 4 + 7 + 10 + 4 = 30 students

c) Mean mark = (mark × frequency)/total frequency

[(6 × 5) + (7 × 4) + (8 × 7) + (9 × 10) + 10 × 4)]/ 30

Mean mark = (30 + 28 + 56 + 90 + 40)/30 = 244/30

Mean mark = 8.1

6 0

3 years ago

The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees. One year, th

Basile [38]

Answer:

The Z-score for that day's temperature is 2.41, and since Z > 2, this temperature is significantly high.

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

If $|Z| > 2$ , the measure X is significantly high(Z > 2) or significantly low(Z < -2).

The mean high temperature on a particular day in January is 31 degrees F and the standard deviation is 8.7 degrees.

This means that $\mu = 31, \sigma = 8.7$

One year, the temperature was 52 degrees F on that day.

This means that $X = 52$ .

What is the Z-score for that day's temperature?

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{52 - 31}{8.7}$

$Z = 2.41$

The Z-score for that day's temperature is 2.41, and since Z > 2, this temperature is significantly high.

8 0

3 years ago