What is 7(5j + 9 + 5j) simplified with commutative properties? WILL GIVE OUT THE BRAINLYIEST

PLEASE HELP! Lines a and b are parallel. What is the measure of angle b?

bija089 [108]

Lines a and b are parallel.
<a = 79
<d = 79
<z = 180 - 79 = 101
<y = 101
<x = 101
<c = 101

4 0

3 years ago

WILL GIVE 11 POINTS How many solutions does the system have?

PSYCHO15rus [73]

There is one answer.

The intersecting points are (0, 5) just for the extra if you graph them both.

7 0

3 years ago

Determine the astm grain size number if 33 grains per square inch are measured at a magnification of 270×

strojnjashka [21]

8.9

The equation for the grain size is expressed as the equality:
Nm(M/100)^2 = 2^(n-1)
where
Nm = number of grains per square inch at magnification M.
M = Magnification
n = ASTM grain size number

Let's solve for n, then substitute the known values and calculate.
Nm(M/100)^2 = 2^(n-1)

log(Nm(M/100)^2) = log(2^(n-1))
log(Nm) + 2*log(M/100) = (n-1) * log(2)
(log(Nm) + 2*log(M/100))/log(2) = n-1
(log(Nm) + 2*log(M/100))/log(2) + 1 = n

(log(33) + 2*log(270/100))/log(2) + 1 = n
(1.51851394 + 2*0.431363764)/0.301029996 + 1 = n
(1.51851394 + 0.862727528)/0.301029996 + 1 = n
2.381241468/0.301029996 + 1 = n
7.910312934 + 1 = n
8.910312934 = n

So the ASTM grain size number is 8.9

If you want to calculate the number of grains per square inch, you'd use the
same formula with M equal to 1. So:
Nm(M/100)^2 = 2^(n-1)
Nm(1/100)^2 = 2^(8.9-1)
Nm(1/10000) = 2^7.9
Nm(1/10000) = 238.8564458
Nm = 2388564.458

Or about 2,400,000 grains per square inch.

7 0

3 years ago

The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio

zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0

3 years ago

0.4612 rounded to the nearest hundredth= what percentage?

Vlad [161]

Question:Round the decimal to the nearest hundredth;then write it as a percentage

Problem: 0.4612=___(<rounded)=___(<percentage)

0.4612 =~ 0,46 (rounded) = 0,46 = 46 (percentage)

7 0

3 years ago