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d1i1m1o1n [39]
3 years ago
6

I need a hug ༼ つ ◕_◕ ༽つ

Mathematics
2 answers:
PIT_PIT [208]3 years ago
7 0

Answer:

*hugs

Step-by-step explanation:

valina [46]3 years ago
4 0

Answer:

<h2>*<em>hugs</em>*</h2>

ur wellcome ^w^

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Adding integers with integer tiles
KIM [24]
C.) -1.
Explanation: 4 + -5 = -1. When you have a negative number plus a positive you subtract. Then take the highest numbers positive or negative symbol. Hope that helps.
3 0
3 years ago
Saturn is 8.867 × 108 miles away from the Sun. Uranus is 1.787 × 109 miles away from the Sun. Approximately how many times farth
GrogVix [38]

Answer:

0.2 times

Step-by-step explanation:

Saturn: 8.867 × 108 = 957.636

Uranus: 1.787 × 109 = 194.783

You take (Uranus: 194.783)

& you divide it by (Saturn: 957.636)

You should get a funky number like 0.203399830415732 but since they said approximately you only need to use the first 2 (don't forget to round)

That would make Uranus 0.2 times as far from the sun

4 0
3 years ago
A television with a 5:3 screen shows an image with a ratio of 5:4 which creates a letter boxed image. What percent of the screen
lions [1.4K]

Answer:

C

Step-by-step explanation:

8 0
3 years ago
I have been failing cause of brainly i could go to a elite school but im on brainly so much i miss assignments :)​
VARVARA [1.3K]

Answer:

ok dont get addicted

Step-by-step explanation:

7 0
3 years ago
Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel
GarryVolchara [31]

Answer:

The unusual X values ​​for this model are: X = 0, 1, 2, 7, 8

Step-by-step explanation:

A binomial random variable X represents the number of successes obtained in a repetition of n Bernoulli-type trials with probability of success p. In this particular case, n = 8, and p = 0.53, therefore, the model is {8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}. So, you have:

P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024

P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215

P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848

P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912

P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695

P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431

P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371

P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442

P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062

The unusual X values ​​for this model are: X = 0, 1, 7, 8

6 0
4 years ago
Read 2 more answers
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