Answer:
RST Is congruent to R’’S’’T’’
Angle R is congruent to angle R prime is congruent to angle R double-prime
TS Is congruent to T’S’ Is congruent to T’’S’’
Step-by-step explanation:
we know that
A reflection and a translation are rigid transformation that produce congruent figures
If two or more figures are congruent, then its corresponding sides and its corresponding angles are congruent
In this problem
Triangles RST, R'S'T and R''S''T'' are congruent
That means
Corresponding sides
RS≅R'S'≅R''S''
ST≅S'T'≅S''T''
RT≅R'T'≅R''T''
Corresponding angles
∠R≅∠R'≅∠R''
∠S≅∠S'≅∠S''
∠T≅∠T'≅∠T''
therefore
RST Is congruent to R’’S’’T’’
Angle R is congruent to angle R prime is congruent to angle R double-prime
TS Is congruent to T’S’ Is congruent to T’’S’’
First, let's write two expressions, letting x= the number of months which have elapsed:
Bill: 120 + 10x
Phil: 150 + 4x
If we set them equal to each other, then solve for x, that will be the number of months where their weights equal each other:
120+10x = 150 + 4x [starting equation]
-120 -4x -120 -4x [ subtract 120 from both sides, and 4x from both sides, to isolate the term with the variable]
<u>6x</u> = <u>30</u> [divide both sides by 5]
5 5
x=6. They will weigh the same in six months.
Answer:
252
Step-by-step explanation:
Split up the shape
First part:
12*12=144
Second part:
((12+(12-6))/2)*24-12
((12+6)/2)*12
18/2*12
9*12 = 108
Total:
108+144 = 252
<u>Plz mark brainliest if this was helpful</u>
So the waiting time for a bus has density f(t)=λe−λtf(t)=λe−λt, where λλ is the rate. To understand the rate, you know that f(t)dtf(t)dt is a probability, so λλ has units of 1/[t]1/[t]. Thus if your bus arrives rr times per hour, the rate would be λ=rλ=r. Since the expectation of an exponential distribution is 1/λ1/λ, the higher your rate, the quicker you'll see a bus, which makes sense.
So define <span><span>X=min(<span>B1</span>,<span>B2</span>)</span><span>X=min(<span>B1</span>,<span>B2</span>)</span></span>, where <span><span>B1</span><span>B1</span></span> is exponential with rate <span>33</span> and <span><span>B2</span><span>B2</span></span> has rate <span>44</span>. It's easy to show the minimum of two independent exponentials is another exponential with rate <span><span><span>λ1</span>+<span>λ2</span></span><span><span>λ1</span>+<span>λ2</span></span></span>. So you want:
<span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span><span>P(X>20 minutes)=P(X>1/3)=1−F(1/3),</span></span>
where <span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span><span>F(t)=1−<span>e<span>−t(<span>λ1</span>+<span>λ2</span>)</span></span></span></span>.
