PLEASE HELP!

Ann travels from town a to town b in 3 hours she then takes another hour to travel from town b to town c the distance between to

abruzzese [7]

add the total time = 3+1 = 4 hours

add total miles = 100+49 = 149

divide 149 by 4= 149/4 = 37.25 miles per hour average

8 0

3 years ago

PLEASE HELP!! I WILL AWARD BRAINLIEST AND 50 POINTS!!!!!!!!!!!!!!!!!!!!!! For the given picture: a) List all of the triangles th

PIT_PIT [208]

Answer:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters. If your teacher just wants one, it would be the middle one. (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB. For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD. For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO. For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB. For OBF, it’s BOC and OCB

Step-by-step explanation:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters. If your teacher just wants one, it would be the middle one. (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB. For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD. For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO. For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB. For OBF, it’s BOC and OCB

6 0

3 years ago

Two friends are collecting stuffed animals. Veronica has 5 more than double the number of stutted

STatiana [176]

v + m = 32 and v = 5 + 2m are the equations that are used to determine m, the number of stuffed animals Mariposa has

Number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23

<h3><u>Solution:</u></h3>

Let "v" be the number of stuffed animals with Veronica

Let "m" be the number of stuffed animals with Mariposa

Given that,

Together, they have 32 stuffed animals

Therefore,

v + m = 32 --------- eqn 1

Veronica has 5 more than double the number of stutted animals as her friend Mariposa

Therefore,

Number of stuffed animals with Veronica = 5 + 2(number of stuffed animals with Mariposa)

v = 5 + 2m ---------- eqn 2

Thus eqn 1 and eqn 2 can be used to determine m, the number of stuffed animals Mariposa has

Let us solve eqn 1 and eqn 2

Substitute eqn 2 in eqn 1

5 + 2m + m = 32

5 + 3m = 32

3m = 32 - 5

3m = 27

Substitute m = 9 in eqn 2

v = 5 + 2(9)

v = 5 + 18

Thus number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23

5 0

3 years ago

The ratio of the weight of Jennifer to that of her friend Laura is 6:11. Laura goes on a diet and loses 12kg and now the ratio o

Inessa05 [86]

Answer:

21kg

Step-by-step explanation:

We will do this.

11x - 12 = 7x

11x - 4x = 7x

4x = 12

4x/4 = 12/4

x = 3

3 x 7 = 21

Laura weighs 21 kg.

3 0

2 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago