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lozanna [386]
3 years ago
6

How do I solve this?

Mathematics
2 answers:
Tems11 [23]3 years ago
7 0

Answer:

x>83/15

Step-by-step explanation:

x-5/2+2/5>2/3

15x-75+12>20 (multiply both sides by 30)

15x-63>20

15x>83

x>83/15

KatRina [158]3 years ago
7 0

Answer:

x>\frac{83}{15}

Step-by-step explanation:

STEP 1: Simplify \frac{x-5}{2} +\frac{2}{5}>\frac{2}{3}

\frac{5x-21}{10}>\frac{2}{3}

STEP 2: Multiply both sides of the equation by 10.

\frac{5x-21}{10}>\frac{2}{3}*(10)

STEP 3: Remove parentheses.

\frac{5x-21}{10}>\frac{2}{3}*(10)

STEP 4: Multiply \frac{2}{3}*(10)

\frac{5x-21}{10}>\frac{20}{3}

STEP 5: Move all terms not containing x to the right side of the inequality.

5x>\frac{83}{3}

STEP 6: Divide each term by 5 and simplify.

x>\frac{83}{15}

STEP 7: The result can be shown in multiple forms.

Inequality Form:

x>\frac{83}{15}

Interval Notation:

(\frac{83}{15},∞)

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A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with σ=0.0
oksian1 [2.3K]
<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

\overline{x}\pm z^* SE                   (1)

, where \overline{x} = sample mean

z* = critical value.

SE = standard error

and SE=\dfrac{\sigma}{\sqrt{n}} , \sigma = population standard deviation.

n= sample size.

As per given , we have

\overline{x}=74.021

\sigma=0.001

n= 15

It is known that ring diameter is normally distributed.

SE=\dfrac{0.001}{\sqrt{15}}=0.000258198889747\approx0.000258199

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 99% two-sided confidence interval on the true mean piston diameter :

74.021\pm (2.576) (0.000258199)     (using (1))

74.021\pm 0.000665120624

74.021\pm 0.000665120624\\\\=(74.021- 0.000665120624,\ 74.021+ 0.000665120624)\\\\=(74.0203348794,\ 74.0216651206)\approx(74.020,\ 74.022)  [Rounded to three decimal places]

∴  A 99% two-sided confidence interval on the true mean piston diameter  = (74.020, 74.022)

By z-table ,

The critical value for 95% confidence  = z*= 1.96

A 95% lower confidence bound on the true mean piston diameter:

74.021- (1.96) (0.000258199)    (using (1))

74.021- 0.00050607004=74.02049393\approx74.020 [Rounded to three decimal places]

∴  A 95% lower confidence bound on the true mean piston diameter= 74.020

7 0
3 years ago
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