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bezimeni [28]
2 years ago
6

I have no idea what it is​

Mathematics
1 answer:
amid [387]2 years ago
3 0

Answer:

answer is option b

2x-3

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yan [13]
There’s no picture or anything.
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Express 59.2475 in words and expanded notation
enyata [817]

Hey, expressing 59.2475 would look like:


Word form: Fifty-nine and two thousand, four hundred seventy-five ten-thousandths.


Expanded Notation Form:

50 + 9 + 0.2 + 0.04 + 0.007 + 0.0005.

6 0
3 years ago
The temperatures at 6:00 am one morning in four cities are given below.
Andre45 [30]

Answer:

-9

Step-by-step explanation:

8 0
2 years ago
After your yearly checkup, the doctor has some bad news and some good news. The bad news is that you tested positive for a serio
Maru [420]

Answer:

0.009804

Step-by-step explanation:

We are given;

probability of testing positive given that you have the disease is 0.99

Also, probability of not testing positive and not having the disease is 0.99

We are also told that it is a rare disease and so strikes only 1 in 1000 people = 0.0001

Let's denote positive test by T+, negative test by T¯, having the disease by D+, not having the disease by D¯.

So, we can now denote all the values in probability we have written earlier.

Thus:

P(T+ | D+) = 0.99

P(T¯ | D¯) = 0.99

P(D+) = 0.0001

Thus, P(D¯) = 1 - P(D+) = 1 - 0.0001 = 0.9999

Now, let's find probability of testing positive;

P(T+) = (P(T+ | D+) × P(D+)) + (P(T+ | D¯) × P(D¯))

Now, (P(T+ | D¯) is not given but by inspection, we can infer from the values given that it is 0.01

Thus;

P(T+) = (0.99 × 0.0001) + (0.01 × 0.9999)

P(T+) = 0.010098

Chances that one has the disease would be gotten from Baye's theorem;

P(D+ | T+) = (P(T+ | D+) × P(D+))/P(T+) = (0.99 × 0.0001)/0.010098 = 0.009804

7 0
3 years ago
Type the correct answer in the box. Write your answer as a reduced fraction, using / for the fraction bar. A six-sided fair die
Schach [20]

Answer:   \bold{\dfrac{125}{1296}}

<u>Step-by-step explanation:</u>

The probability of NOT getting a 4 means the probability of getting a 1, 2, 3, 5, or 6 which is 5 out of 6 = \dfrac{5}{6}

The probability of getting a 4 is 1 out of 6 = \dfrac{1}{6}

The probability of getting a 4 only on the last roll is:

NOT 4    and     NOT 4    and    NOT 4   and     Yes 4

    \dfrac{5}{6}           x            \dfrac{5}{6}            x          \dfrac{5}{6}           x           \dfrac{1}{6}      =      \dfrac{5^3}{6^4}= \dfrac{125}{1296}

3 0
3 years ago
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