Answer:Thanks
Step-by-step explanation:
The graph will be as shown in attached figure
Answer:
The diagonals of an isosceles trapezoid are congruent.
The bases of a trapezoid are parallel.
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
The answer is B) <span>Stacy’s employer is withholding too little for state income tax.
Stacy's state income tax withholding should be:
163.88 x 19% = $31.14, which is less than $24.58</span>
