Answer:
Question 7:
∠L = 124°
∠M = 124°
∠J = 118°
Question 8:
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15:
m∠G = 110°
Question 16:
∠G = 60°
Question 17:
∠G = 80°
Question 18:
∠G = 70°
Step-by-step explanation:
The angles can be solving using Symmetry.
Question 7.
The sum of interior angles in an isosceles trapezoid is 360°, and because it is an isosceles trapezoid
∠K = ∠J = 118°
∠L = ∠M
∠K+∠J+∠L +∠M = 360°
236° + 2 ∠L = 360°
Therefore,
∠L = 124°
∠M = 124°
∠J = 118°
Question 8.
In a similar fashion,
∠Q+∠T+∠S +∠R = 360°
and
∠R = ∠S = 82°
∠Q = ∠T
∠Q+∠T + 164° = 360°
2∠Q + 164° = 360°
2∠Q = 196°
∠Q = ∠T =98°.
Therefore,
∠Q = 98°
∠T = 98°
∠R = 82°
Question 15.
The sum of interior angles of a kite is 360°.
∠E + ∠G + ∠H + ∠F = 360°
Because the kite is symmetrical
∠E = ∠G.
And since all the angles sum to 360°, we have
∠E +∠G + 100° +40° = 360°
2∠E = 140° = 360°
∠E = 110° = ∠G.
Therefore,
m∠G = 110°
Question 16.
The angles
∠E = ∠G,
and since all the interior angles sum to 360°,
∠E + ∠G + ∠F +∠H = 360°
∠E + ∠G + 150 + 90 = 360°
∠E + ∠G = 120 °
∠E = 60° = ∠G
therefore,
∠G = 60°
Question 17.
The shape is a kite; therefore,
∠H = ∠F = 110°
and
∠H + ∠F + ∠E +∠G = 360°
220° + 60° + ∠G = 360°,
therefore,
∠G = 80°
Question 18.
The shape is a kite; therefore,
∠F = ∠H = 90°
and
∠F +∠H + ∠E + ∠G = 360°
180° + 110° + ∠G = 360°
therefore,
∠G = 70°.
Answer:
slope: 5/7
0.714 mile per hour
Step-by-step explanation:
slope: find the point where the line is right on the line graph, put y/x which is 10/14 make the fraction smaller which is 10/14 = 5/7(I divide the denominator and nominator by 2)
Mile per hour: 5/7 cross multiply with x/1, make into an equation:
7x=5 x 1
7x = 5
x = 0.714
Step-by-step explanation:
=2* ( l+b )
=2* ( 11.37+9.245 )
=2* 20.595
=41.19
Answer:
So Philip made 5 bracelets and 4 necklaces.
Step-by-step explanation:
Let x = number of bracelets and y = number of necklaces.
Since we have a total of 9 bracelets and necklaces,
x + y = 9 (1)
Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.
So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,
8x + 20y = 120 (2)
Simplifying it we have
2x + 5y = 30 (3).
Writing equations (1) and (3) in matrix form, we have
![\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%5C%5C30%5Cend%7Barray%7D%5Cright%5D)
Using Cramer's rule to solve for x and y,
![x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=x%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%261%5C%5C30%265%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)
x = (45 - 30) ÷ (5 - 2)
x = 15 ÷ 3
x = 5
![y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=y%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%269%5C%5C2%2630%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)
y = (30 - 18) ÷ (5 - 2)
y = 12 ÷ 3
y = 4
So Philip made 5 bracelets and 4 necklaces.
<span>(a) This is a binomial
experiment since there are only two possible results for each data point: a flight is either on time (p = 80% = 0.8) or late (q = 1 - p = 1 - 0.8 = 0.2).
(b) Using the formula:</span><span>
P(r out of n) = (nCr)(p^r)(q^(n-r)), where n = 10 flights, r = the number of flights that arrive on time:
P(7/10) = (10C7)(0.8)^7 (0.2)^(10 - 7) = 0.2013
Therefore, there is a 0.2013 chance that exactly 7 of 10 flights will arrive on time.
(c) Fewer
than 7 flights are on time means that we must add up the probabilities for P(0/10) up to P(6/10).
Following the same formula (this can be done using a summation on a calculator, or using Excel, to make things faster):
P(0/10) + P(1/10) + ... + P(6/10) = 0.1209
This means that there is a 0.1209 chance that less than 7 flights will be on time.
(d) The probability that at least 7 flights are on time is the exact opposite of part (c), where less than 7 flights are on time. So instead of calculating each formula from scratch, we can simply subtract the answer in part (c) from 1.
1 - 0.1209 = 0.8791.
So there is a 0.8791 chance that at least 7 flights arrive on time.
(e) For this, we must add up P(5/10) + P(6/10) + P(7/10), which gives us
0.0264 + 0.0881 + 0.2013 = 0.3158, so the probability that between 5 to 7 flights arrive on time is 0.3158.
</span>
