Answer:
the probability of a client owning stocks OR bonds is 0.825
Step-by-step explanation:
The computation of the probability of a client owning stocks OR bonds is given below:
= P(stocks) + P(bonds) - P(stocks and bonds)
= 0.60 + 0.50 - (0.55 × 0.50)
= 0.60 + 0.50 - 0.275
= 0.825
Hence, the probability of a client owning stocks OR bonds is 0.825
please help, thanks :)
Answer:
D :)
Step-by-step explanation:
Solve the system by substitution.
-9x – 2y = -44
8x – 3 = y
Answer:
x = 2, y = 13
(2, 13)
Step-by-step explanation:
hope this helps!
please help asap!!!!
Answer:
z = 25
Step-by-step explanation:
adjacent angles in a rhombus are supplementary
3z - 15 + 7z - 55 = 180
10z - 70 = 180
10z = 250
z = 25
In ΔHIJ, the measure of ∠J=90°, the measure of ∠I=60°, and JH = 80 feet. Find the length of IJ to the nearest foot.
Answer:
46.2
Step-by-step explanation:
46.188= 46.2
Help plz:)))I’ll mark u Brainliest
Answer:
Scale factor = 1.5
Step-by-step explanation:
Formula for the scale factor,
Scale factor = [tex]\frac{\text{Measure of one side of the Image}}{\text{Measure of one side of the preimage}}[/tex]
In this question, image is triangle Y and preimage is triangle X.
Scale factor of X to Y = [tex]\frac{15}{10}[/tex]
= 1.5
Therefore, scale factor = 1.5 is the answer.
Solve the triangle given that a=19 b=16, c=11.
Answer:
The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.
Step-by-step explanation:
From statement we know all sides of the triangle ([tex]a[/tex], [tex]b[/tex], [tex]c[/tex]), but all angles are unknown ([tex]A[/tex], [tex]B[/tex], [tex]C[/tex]). (Please notice that angles with upper case letters represent the angle opposite to the side with the same letter but in lower case) From Geometry it is given that sum of internal angles of triangles equal 180º, we can obtain the missing information by using Law of Cosine twice and this property mentioned above.
If we know that [tex]a = 19[/tex], [tex]b = 16[/tex] and [tex]c = 11[/tex], then the missing angles are, respectively:
Angle A
[tex]a^{2} = b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A[/tex] (1)
[tex]A = \cos^{-1}\left(\frac{b^{2}+c^{2}-a^{2}}{2\cdot b\cdot c} \right)[/tex]
[tex]A = \cos^{-1}\left[\frac{16^{2}+11^{2}-19^{2}}{2\cdot (16)\cdot (11)} \right][/tex]
[tex]A \approx 87.395^{\circ}[/tex]
Angle B
[tex]b^{2} = a^{2}+c^{2}-2\cdot a\cdot c \cdot \cos B[/tex] (2)
[tex]B = \cos^{-1}\left(\frac{a^{2}+c^{2}-b^{2}}{2\cdot a\cdot c} \right)[/tex]
[tex]B = \cos^{-1}\left[\frac{19^{2}+11^{2}-16^{2}}{2\cdot (19)\cdot (11)} \right][/tex]
[tex]B\approx 57.271^{\circ}[/tex]
Angle C
[tex]C = 180^{\circ}-A-B[/tex]
[tex]C = 180^{\circ}-87.395^{\circ}-57.271^{\circ}[/tex]
[tex]C = 35.334^{\circ}[/tex]
The angles of the triangle are approximately 87.395º, 57.271º and 35.334º.
Need help #2. The answer is shown, but I don’t know how to get to the answer. Please teach and show steps.
Answer:
please refer to the above attachment
(D) is the right answer.