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Alenkinab [10]
3 years ago
5

7) 1 is what percent of 52.6?

Mathematics
2 answers:
Olin [163]3 years ago
6 0

Answer:

Step-by-step explanation:

0.0190

katovenus [111]3 years ago
3 0

Answer:

1 is what percent of 52.6? = 1.9011406844106.

Step-by-step explanation:

urm google it that is what i did

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Which is the domain of the function f(x)=-5/6[3/5]^*
Snezhnost [94]

Answer:

5/63/5xxx equal to 0

Step-by-step explanation:

^^^^

6 0
2 years ago
Urgent help!<br> Simplify (x − 4)(4x^2 + x − 6).
m_a_m_a [10]

Answer:

2X-1

Step-by-step explanation:

as far as i got

6 0
3 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
3 years ago
Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
My name is Ann [436]

Answer:

a. The value of alpha is 3.014 and the value of beta is 12.442

b. The probability that data transfer time exceeds 50ms is 0.238

c. The probability that data transfer time is between 50 and 75 ms is 0.176

Step-by-step explanation:

a. According to the given data we have that the mean and standard deviation of the random variable X are 37.5 ms and 21.6.

Therefore, E(X)=37.5 and V(X)=(21.6)∧2  

To calculate alpha we would have to use the following formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To calculate beta we would have to use the following formula:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. E(X)=37.5 and V(X)=(21.6)∧2  

Therefore, P(X>50)=1−P(X≤50)

Hence, To calculate the probability that data transfer time exceeds 50ms we use the following formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The probability that data transfer time exceeds 50ms is 0.238

c. E(X)=37.5 and V(X)=(21.6)∧2  

​Therefore, P(50<X<75)=P(X<75)−P(X<50)  

Hence, To calculate the probability that data transfer time is between 50 and 75 ms we use the following formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time is between 50 and 75 ms is 0.176

6 0
3 years ago
Una funzione inversamente proporzionale è rappresentata da?
puteri [66]
Supponiamo che due variabili x e y siano inversamente proporzionali; allora possiamo rappresentarli come x ∝ 1 / y. Ciò significa che se il valore di x aumenta, il valore di y diminuisce e viceversa.


Suppose two variables x and y are inversely proportional; then we can represent them as x ∝ 1/y. That means, if the value of x increases, then the value of y decreases and vice versa.

3 0
2 years ago
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