BRAINLIEST a. 7 sqrt of 2 b. 14 c. 7 d. 7 sqrt of 3

Solve the equation: b - 1.6 ÷ 4 = - 3 b = _____

notka56 [123]

B=-2.6
Hope this helps :)

8 0

3 years ago

Read 2 more answers

Please help!! Which of the following is equal to the rational expression when x ≠ 2 or -4? 5(x-2)/(x-2)(x+4)

Stels [109]

Answer:

5 / (x+4) x ≠2 x≠-4

Step-by-step explanation:

5(x-2)/(x-2)(x+4)

The denominator cannot be zero so x ≠2 x≠-4

Cancel like terms in the numerator and denominator

5 / (x+4) x ≠2 x≠-4

6 0

3 years ago

Find two consecutive even integers such that five times the smaller integer is ten more than three times the larger integer.

quester [9]

5n = 3(n + 2) + 10
5n = 3n + 16
12n = 16
n = 8

8 is smaller integer and 10 is the larger.

8 0

3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

4 years ago

The time that a randomly selected individual waits for an elevator in an office building has a uniform distribution over the int

rewona [7]

Answer:

The mean of the sampling distribution of x is 0.5 and the standard deviation is 0.083.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .