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garri49 [273]
3 years ago
5

43 tens times 3 tens equal how many hundreds

Mathematics
2 answers:
velikii [3]3 years ago
7 0
<span>129 hundreds, because if you multiply 43×3 that should give you 129 so </span>t<span>hat's why it is 129 hundreds.</span>
Readme [11.4K]3 years ago
6 0
I did this years ago...
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5) Two machines M1, M2 are used to manufacture resistors with a design
Basile [38]

Answer:

Since M1 has the higher probability of being in the desired range, we choose M1.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.

So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.

For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.

M1:

Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that \mu = 1050, \sigma = 100

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

Z = \frac{X - \mu}{\sigma}

Z = \frac{1100 - 1050}{100}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 900

Z = \frac{X - \mu}{\sigma}

Z = \frac{900 - 1050}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.6915 - 0.0668 = 0.6247.

M1 has a 62.47% probability of being in the desired range.

M2:

M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that \mu = 1000, \sigma = 120

X = 1100

Z = \frac{X - \mu}{\sigma}

Z = \frac{1100 - 1000}{120}

Z = 0.83

Z = 0.83 has a pvalue of 0.7967.

X = 900

Z = \frac{X - \mu}{\sigma}

Z = \frac{900 - 1000}{120}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

0.7967 - 0.2033 = 0.5934

M2 has a 59.34% probability of being in the desired range.

Since M1 has the higher probability of being in the desired range, we choose M1.

8 0
3 years ago
The sum of three numbers is 8. The third is 9
Ierofanga [76]

Answer:

→<u> </u><u>First</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>1</u>

→<u> </u><u>Second</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>2</u>

→<u> </u><u>Third</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>5</u>

Step-by-step explanation:

• let numbers be x, y and z

{ \tt{z = 8y - 9 -  -  - (eqn \: 1)}} \\  \\ { \tt{10x = 8y - 7 -  -  - (eqn \: 2)}} \\  \\ { \tt{x + y + z = 8 -  -  - (eqn \: 3)}}

• from eqn 2, make x the subject:

{ \tt{x =  \frac{8y - 7}{10} }} \\

• substitute all variables in eqn 3:

{ \tt{ \frac{8y - 7}{10}  + y + 8y - 9 = 8}} \\  \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\  \\ { \tt{98y = 177}} \\  \\ { \boxed{ \tt{ \: y = 1.8}}}

• find z

{ \tt{z = 8y - 9}} \\  \\ { \tt{z = 8(1.8) - 9}} \\  \\ { \tt{z = 14.4 - 9}} \\  \\ { \boxed{ \tt{ \: z = 5.4 \: }}}

• find x:

{ \tt{x =  \frac{8(1.8) - 7}{10} }} \\  \\ { \tt{x =  \frac{7.4}{10} }} \\  \\ { \boxed{ \tt{ \: x = 0.74 \: }}}

Rounding to nearest value:

{ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{  \rm{z = 5}}}

7 0
3 years ago
If three points are collinear, they are also coplanar
e-lub [12.9K]

My explanation is attached below.

6 0
3 years ago
Read 2 more answers
Multiply 3 by the number of days in a week. Subtract 12 and write your
k0ka [10]

Answer:

9?

Step-by-step explanation:

its c a l c u l a t o r timeee.

-3 * 7= 21

21 - 12= 9

so now how do you put 9 in the thousands place it's just in the ones place which won't make sense

8 0
3 years ago
Read 2 more answers
The table displays the frequency of scores for a calculus class on an exam. The mean of the exam scores is 3.5. Score 1 2 3 4 5
vladimir1956 [14]

Answer:

f=11

Mode=4

Median=4

Explanation:

We are given that

a.Mean of the exam score,\bar x=3.5

Score(x)   frequency   C.F

1                  1                 1

2                  3                4

3                  f                 15(4+f=4+11)

4                  13               28

5                   4               32

\sum f_i=1+3+f+13+4=21+f

\sum f_ix_i=1(1)+2(3)+3(f)+4(13)+5(4)=1+6+3f+52+20=79+3f

(\bar x)=\frac{\sum f_ix_i}{\sum f_i}

Using the formula

3.5=\frac{79+3f}{21+f}

73.5+3.5f=79+3f

3.5f-3f=79-73.5

0.5f=5.5

f=\frac{5.5}{0.5}=11

b.Mode:The number which is repeat most times .

4 repeat most times

Hence, mode of all exam scores=4

N=32

N is even

Median=\frac{(\frac{n}{2})^{th}+(\frac{n}{2}+1)^{th}}{2}

Median=\frac{16th+17th}{2}=\frac{4+4}{2}=\frac{8}{2}=4

7 0
3 years ago
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