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DIA [1.3K]
3 years ago
14

Please help me please!!​

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

mok

Step-by-step explanation:

mok

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48

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3 years ago
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%of 25 is 14.<br><br> what percent of 20 is 11?<br><br> 15 is
ikadub [295]
The first question:

'is' in mathematics means equal
'of' in mathematics means multiplication

\frac{x}{100} \times 25 = 14\\\\\frac{x}{4} = 14\\\\\boxed{\bf{x=56}}

56% of 25 is 14.

And the second question:

'is' in mathematics means equal
'of' in mathematics means multiplication

\frac{x}{100} \times 20 = 11\\\\\frac{x}{5} = 11\\\\\boxed{\bf{x = 55}}

55% of 20 is 11<span>.</span>
4 0
4 years ago
Helllllllllllllllllppppppppppppppppppp
Gennadij [26K]
4 1/10 then 1/4 then -4 then -4 3/4 hope that helps
3 0
3 years ago
Read 2 more answers
Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to
levacccp [35]
P = 0.11
1-p = .89
n = 200

mean = n*p = 22
std dev = n*sqrt(p(1-p)/n) = 4.425

find z values for standard normal distribution
Z= (30-22)/4.425 = 1/808
look up table to find P(Z less than 1.808) which is same P(x less than 30)

for 2nd part, you need 2 Z values
Z1 = (25-22) / 4.425 = 0.678
Z2 = (15-22) / 4.425 = -1.58

P(-1.58 less than Z less than 0.678) = P(Z less than .678) - P(Z less than - 1.58)

5 0
4 years ago
A twelve-sided die with sides labeled through will be rolled once. Each number is equally likely to be rolled. What is the proba
FromTheMoon [43]

<em>Complete Question:</em>

<em>A twelve-sided die with sides labeled 1 through 12 will be rolled once. Each number is equally likely to be rolled, what is the probability of rolling a number greater than 10?</em>

Answer:

P(T) = \frac{1}{6}

Step-by-step explanation:

Given

Number of Sides = 12

Required

Probability of obtaining a side greater than 10

We start by listing out the sample space;

S = \{1,2,3,4,5,6,7,8,9,10,11,12\}

n(S) = 12

Next, we list out digits greater than 10; Represent this with T

T = \{11,12\}

n(T) = 2

Probability of T is calculated as follows;

P(T) = \frac{n(T)}{n(S)}

P(T) = \frac{2}{12}

Divide the numerator and denominator by 2

P(T) = \frac{1}{6}

<em>Hence, the required probability is </em>\frac{1}{6}<em />

8 0
3 years ago
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