$10.75 x 8 = $86 x 4(weeks) = <em>$344</em>
Well, he only needs 5 gallons of beverages. Because he buys two containers of each, however, that means he bought 320 gallons total (I would say). How I got my answer:
There are 8 pints in a gallon (8 x 2)
There are 4 quarts in a gallon (4 x 2)
There are 16 cups in a gallon (16 x 2)
There are 128 ounces in a gallon (128 x 2)
With that being said, we can add these to find the total amount of gallons of beverages Julius bought. 16 + 8 + 8 + 32 + 256 = 320.
Hopefully this helps!
Answer:
PG ≅ SG (Given)
PT ≅ ST (Given)
GT = GT (Common)
∴ ∠GPT ≅ ∠GST (SSS Congruency Axiom)
Step-by-step explanation:
<u>Given</u>: PG ≅ SG and PT ≅ ST
<u>To Prove</u>: ∠GPT ≅ ∠GST
<u>Proof</u>: PG ≅ SG (Given)
PT ≅ ST (Given)
GT = GT (Common)
∴ ∠GPT ≅ ∠GST (SSS Congruency Axiom).
<u>SSS Congruency Axiom</u>: If three pairs of sides of two triangles are equal in length, then the triangles are congruent.
<u>Congruence</u>: Two sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of rigid motions, namely a translation, a rotation, and a reflection. This means that either object can be repositioned and reflected (but not resized) so as to coincide precisely with the other object. Two triangles are congruent if their corresponding sides are equal in length, and their corresponding angles are equal in measure.
Answer:16???
Step-by-step explanation:
Answer:
a) <u>0.4647</u>
b) <u>24.6 secs</u>
Step-by-step explanation:
Let T be interval between two successive barges
t(t) = λe^λt where t > 0
The mean of the exponential
E(T) = 1/λ
E(T) = 8
1/λ = 8
λ = 1/8
∴ t(t) = 1/8×e^-t/8 [ t > 0]
Now the probability we need
p[T<5] = ₀∫⁵ t(t) dt
=₀∫⁵ 1/8×e^-t/8 dt
= 1/8 ₀∫⁵ e^-t/8 dt
= 1/8 [ (e^-t/8) / -1/8 ]₀⁵
= - [ e^-t/8]₀⁵
= - [ e^-5/8 - 1 ]
= 1 - e^-5/8 = <u>0.4647</u>
Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>
<u></u>
b)
Now we find t such that;
p[T>t] = 0.95
so
t_∫¹⁰ t(x) dx = 0.95
t_∫¹⁰ 1/8×e^-x/8 = 0.95
1/8 t_∫¹⁰ e^-x/8 dx = 0.95
1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t = 0.95
- [ e^-x/8]¹⁰_t = 0.96
- [ 0 - e^-t/8 ] = 0.95
e^-t/8 = 0.95
take log of both sides
log (e^-t/8) = log (0.95)
-t/8 = In(0.95)
-t/8 = -0.0513
t = 8 × 0.0513
t = 0.4104 (min)
so we convert to seconds
t = 0.4104 × 60
t = <u>24.6 secs</u>
Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>
