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Bogdan [553]
3 years ago
7

Find w(4) for the function w( x ) = 3x + 7 A. 41 B. 33 C. 14 D. 19

Mathematics
1 answer:
leonid [27]3 years ago
4 0
The answer is letter b
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Which number represents "nine and eighty-one thousandths?"
Lemur [1.5K]

Answer:

9.081

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​ However, the
kumpel [21]

Answer:

Check the explanation

Step-by-step explanation:

Let X denotes steel ball and Y denotes diamond

\bar{x_1} = 1/9( 50+57+......+51+53)

=530/9

=58.89

\bar{x_2}= 1/9( 52+ 56+....+ 51+ 56)

=543/9

=60.33

difference = d =(60.33- 58.89)

=1.44

s^2=1/n\sum xi^2 - n/(n-1)\bar{x}^2

s12 = 1/9( 502+572+......+512+532) -9/8 (58.89)2

=31686/8 - 9/8( 3468.03)

=3960.75 - 3901.53

=59.22

s1 = 7.69

s22 = 1/9( 522+ 562+....+ 512+ 562) -9/8 (60.33)2

=33295/8 - 9/8 (3640.11)

=4161.875 - 4095.12

=66.75

s2 =8.17

sample standard deviation for difference is

s=\sqrt{[(n1-1)s_1^2+ (n2-1)s_2^2]/(n1+n2-2)}

 = \sqrt{[(9-1)*59.22+ (9-1)*66.75]/(9+9-2)}

= \sqrt{1007.76/16}

=7.93

sd = s*\sqrt{(1/n1)+(1/n2)}

=7.93*\sqrt{(1/9)+(1/9)}

=7.93* 0.47

=3.74

For 95% confidence level Z (\alpha /2) =1.96

confidence interval is

d\pm Z(\alpha /2)*s_d

=(1.44 - 1.96* 3.75 , 1.44+1.96* 3.75)

=(1.44 - 7.35 , 1.44 + 7.35)

=(-2.31, 8.79)

There is sufficient evidence to conclude that the two indenters produce different hardness readings.

3 0
3 years ago
Use math drawings to make pictures equal
gavmur [86]
Well, what are the pictures??
5 0
3 years ago
A number sentence is said to be true if both numerical expressions are
Igoryamba
If both expressions are equivalent (that is when both evaluate to the same number)
6 0
3 years ago
Shortly after their arrival, Europeans began introducing pigs to the Americas as a source of food. Some escaped, and others were
natka813 [3]

Answer: About 800 thousand

The more accurate value is 807,528 but this is also an approximation.

=======================================================

Work Shown:

t = \text{number of years since the year 2000}

P_0 = \text{population (in millions) in the year 2000}

P = P_0(1.2)^t\\\\5 = P_0(1.2)^{10}\\\\5 \approx P_0*6.1917364224\\\\P_0 \approx \frac{5}{6.1917364224}\\\\P_0 \approx 0.80752791444922\\\\P_0 \approx 0.807528\\\\

That's the rough population of wild pigs (in millions) for the year 2000.

Multiply by 10^6 to get it in terms of units instead.

0.807528\times10^6 = 807,528

There were roughly 800 thousand wild pigs in the year 2000.

7 0
2 years ago
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