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PIT_PIT [208]
3 years ago
5

4.56 is greater than or equal to 1.9d solve for d

Mathematics
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

2,4 ≥ d

Step-by-step explanation:

\frac{4,56}{1,9} ≥ \frac{1,9d}{1,9} \\ \\ 2,4 ≥ d

I am joyous to assist you anytime.

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The circumference of C is 72 cm. What is the length of (the minor arc)? A. 90 cm B. 10 cm C. 45 cm D. 8 cm E. 9 cm F. 7.2 cm
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Answer:

e. 9cm

Step-by-step explanation:

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Step-by-step explanation:

4 0
3 years ago
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Is the one i picked right​
motikmotik
The arithmetic sequence formula should help you
An= A1+ (n-1)d or An= a + d(n−1)
An= the term that you want to find.
A1 = first term in the list of ordered numbers
n = the term position (ex: for 5th term, n = 5 )
d = common difference of any pair of consecutive or adjacent numbers
So, looking the 4 choices you have the only one that follow the sequence of the formula is C
I believe the one you choose is Fibonacci sequence.
There different number of sequences:
1- arithmetic sequence
2- geometry sequence
3- Fibonacci sequence
Each of that sequences have their own formula.
So, the answer it might be C
Hope that help you




The formula for the Arithmetic is:
An = A1 + F x ( n- 1)
I believe answer is C
Because the one you chose is
7 0
3 years ago
Estimate the product of 79.623 times 4.83
r-ruslan [8.4K]
It would be 384.58, or 384.57909
4 0
3 years ago
Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv
Dahasolnce [82]

Answer and Explanation:

Given : Five males with an​ X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the​ X-linked genetic disorder.

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029

\sum P(X)=1

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

\mu=\sum xP(x)=2.5

The standard deviation of the random sample is

Variance=\sigma^2

\sigma^2=\sum x^2P(x)-\mu^2

\sigma^2=7.436-(2.5)^2

\sigma^2=7.436-6.25

\sigma^2=1.186

\sigma=\sqrt{1.186}

\sigma=1.08

Therefore, The mean is 2.5 and the standard deviation is 1.08.

5 0
3 years ago
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