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Jlenok [28]
3 years ago
5

Emir made $182 in interest by placing $700 in a savings account with simple interest for 2 years. What was the interest rate?

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
6 0

Answer:

7.69230%

R=I/P×T

R=$182/$700×2

R=182/1400

R=91/700

=7.69230%

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A study has shown that the daily amount of milk produced by a dairy cow is normally distributed, with a mean of 6.2 gallons and
V125BC [204]

Answer:

5.5 gallons to 6.9 gallons

Step-by-step explanation:

Daily a cow

produces a mean of 6.2 gallons (average

of 6.2 gallons, meaning they already

found the mean, "averaged" it out for

you) with a deviation of 7 gallons ("give

or take" .7 gallons). On any given day

(throw these words out) 68% of the cows

(don't need this information to answer

the question either, thrown in to confuse

you) will produce an amount of milk

within which of the following ranges?

Just subtract the deviation ("give or

take number").7 gallons from the mean

(average (already determined)) 6.2

gallons which is 5.5 gallons THEN add

the deviation ("give or take number").

7 gallons to the mean (average (already

determined)) 6.2 gallons which is 6.9

gallons. Your answer is A 5.5 gallons to

6.9 gallons. You don't even have to go

crazy on the math on this question you

can rule them all out but A immediately

with subtracting .7 from 6.2.

hope it helps!

4 0
3 years ago
Read 2 more answers
Find the value of x if h(x) = -25 in the function h(X) = 1/5x + 10
Dmitry_Shevchenko [17]

Answer:

x = -175

Step-by-step explanation:

Step 1: Define

h(x) = -25

h(x) = 1/5x + 10

Step 2: Substitute and Evaluate

-25 = 1/5x + 10

-35 = 1/5x

x = -175

3 0
3 years ago
Read 2 more answers
g A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of pa
adoni [48]

Complete Question

A milling process has an upper specification of 1.68 millimeters and a lower specification of 1.52 millimeters. A sample of parts had a mean of 1.6 millimeters with a standard deviation of 0.03 millimeters. what standard deviation will be needed to achieve a process capability index f 2.0?

Answer:

The value required is  \sigma =  0.0133

Step-by-step explanation:

From the question we are told that

   The upper specification is  USL  =  1.68 \ mm

    The lower specification is  LSL  = 1.52  \  mm

     The sample mean is  \mu =  1.6 \  mm

     The standard deviation is  \sigma =  0.03 \ mm

Generally the capability index in mathematically represented as

             Cpk  =  min[ \frac{USL -  \mu }{ 3 *  \sigma }  ,  \frac{\mu - LSL }{ 3 *  \sigma } ]

Now what min means is that the value of  CPk is the minimum between the value is the bracket

          substituting value given in the question

           Cpk  =  min[ \frac{1.68 -  1.6 }{ 3 *  0.03 }  ,  \frac{1.60 -  1.52 }{ 3 *  0.03} ]

=>      Cpk  =  min[ 0.88 , 0.88  ]

So

         Cpk  = 0.88

Now from the question we are asked to evaluated the value of  standard deviation that will produce a  capability index of 2

Now let assuming that

         \frac{\mu - LSL  }{ 3 *  \sigma } =  2

So

         \frac{ 1.60 -  1.52  }{ 3 *  \sigma } =  2

=>    0.08 = 6 \sigma

=>     \sigma =  0.0133

So

        \frac{ 1.68  - 1.60 }{ 3 *  0.0133 }

=>      2

Hence

      Cpk  =  min[ 2, 2 ]

So

    Cpk  = 2

So    \sigma =  0.0133 is  the value of standard deviation required

3 0
3 years ago
I will give u brainlyist!! pls help
Karolina [17]

Answer:

i believe C

Step-by-step explanation:

5 0
3 years ago
Help me solve this questions with steps please
Flauer [41]

Answer:

B

Step-by-step explanation:

Since the triangle is right use the tangent ratio to find x

tan22° = \frac{opposite}{adjacent} = \frac{203}{x}

Multiply both sides by x

x × tan22° = 203 ( divide both sides by tan22° )

x = \frac{203}{tan22} ≈ 502.5 m → B

7 0
3 years ago
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