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hammer [34]
4 years ago
6

Control charts on X and S are to be maintained on the torque reading of a bearing used in a wingflap actuator assembly. Samples

of size n-8 are to be used, and we know from past experience that when the process is in control, bearing torque has a normal distribution with mean μ-: 80 inch-pounds and standard deviation σ center line and control limits for these control charts -12 inch-pounds. Find the center line and control limits for these control charts.
Business
1 answer:
Marina CMI [18]4 years ago
5 0

Answer:

Central line will be mean =80 inch pound

Upper control lmit = 80+3*sigma =80+3*12=116

Lower control lmit = 80-3*sigma =80-3*12=44

Explanation:

Central line will be mean =80 inch pound

Upper control lmit = 80+3*sigma =80+3*12=116

Lower control lmit = 80-3*sigma =80-3*12=44

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the capacity to provide future services or benefits.

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if it was determined that the movement of exchange rates was not related to previous exchange rate values, this implies that a i
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The Delphi method, forecast by analogy, growth curves, extrapolation, and horizon scanning are all widely used tools for technology forecasting.

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8 0
1 year ago
Universal Foods issued 10% bonds, dated January 1, with a face amount of $260 million on January 1, 2018. The bonds mature on De
kondaur [170]

Answer:

The bonds were issued at $220,879,628.13

This is lower than the face value to compensate for the lower coupon payment.

cash               220,879,628.13   debit

discount on BP  39,120,371.87   debit

   bonds payable      260,000,000 credit

--to record the issuance of the bonds--

Interest expense 13,252,777.69 debit

Discoun on BP               252,777.69 credit

 cash          13,000,000      credit

--to record the first interest payment--

Interest expense 13,267,944.35 debit

        Discount on BP                267,944.35 credit

 Cash          13,000,000     credit

--to record second interest payment--

Interest expense 13,539,156.67 debit

Discount on BP              539,156.67 credit

cash                   13,000,000.00 credit

--to record Dec 31st, 2025 payment--

Explanation:

To determinate the price we will solve for the present value of the coupon payment and maturity at the market rate of %12

C \times \frac{1-(1+r)^{-time} }{rate} = PV\\

Coupon payment:

260,000,000 x 10% x 1/2 =13,000,000.000

time 20 years x 2 payment per year 40

yield to maturity  12% / 2 = 6%

13000000 \times \frac{1-(1+0.06)^{-40} }{0.06} = PV\\

PV $195,601,859.3298

\frac{Maturity}{(1 + rate)^{time} } = PV  

Maturity   260,000,000.00

time   40.00

rate  0.06

\frac{260000000}{(1 + 0.06)^{40} } = PV  

PV   25,277,768.80

PV c $195,601,859.3298

PV m  $25,277,768.8042

Total $220,879,628.1340

For the journal entries, we will multiply this current market price of the bonds by the market rate (YTM) the difference between this and the actual cash obligation generate by the bond is the amortization of the discount.

<u>first interest payment </u>

$220,879,628.13 x 6% = 13,252,777.69

less actual cash outlay:  13,000,000

amortization                          252,777.69

<u>second interest payment</u>

($220,879,628.13- $252,777.69) x 6% = 13,267,944.35

less actual cash outlay:                      <u>     13,000,000.00</u>

amortization                                                   267,944.35

December 31st, 2025:

This will be payment 14th

after building the schedule until that date we got:

8 0
3 years ago
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