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3 years ago

Please help me find the answers to these

Mathematics

2 answers:

maksim [4K]3 years ago

4 0

A. 3 miles
B. 9 miles
C. 5 hours
D. 7 hours
E. 12 miles
F. 11 hours
G. 3 miles
H. 6 hours
i might be wrong on some but i’m pretty sure it’s correct

AnnZ [28]3 years ago

3 0

Answer:

A. 12 miles

B. 9 miles

C. 2hours 30mins

D. 7hours

sry thats all i got

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Can someone please help me?

kotykmax [81]

Answer:part a=4.6 part b=2.6

Step-by-step explanation:part a= first make sure to know which direction is which next you would end up making a triangle with the legs as 1.7 and 4.3 and use pythagorean theorem a^2+b^2=c^2 so 1.7^2 + 4.3^2= c^2 1.7^2=2.89 4.3^2=18.49 2.89+18.49=21.38 so now you square root 21.38 to get 4.62385 so it equals 4.6

part b = is basically the same thing except they gave you 3.1 as c and 1.7 as a so you would put it as c^2 - a^2=b^2 so 9.61 - 2.89= 6.72 then square root it to get 2.592296 which means it equals 2.6

3 0

3 years ago

Need this answer as fast as possible

Greeley [361]

D, negatives of A and B cancel each other out.

6 0

3 years ago

Read 2 more answers

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

WARRIOR [948]

Answer:

(a) The probability that more than 180 will take your free sample is 0.1056.

(b) The probability that fewer than 200 will take your free sample is 0.9997.

(c) The probability that a customer will take a free sample and buy the product is 0.2184.

(d) The probability that between 60 and 80 customers will take the free sample and buy the product is 0.8005.

Step-by-step explanation:

We are given that about 56% of all customers will take free samples. Furthermore, of those who take the free samples, about 39% will buy what they have sampled.

The day you were offering free samples, 303 customers passed by your counter.

Firstly, we will check that it is appropriate to use the normal approximation to the binomial, that is;

Is np > 5 and n(1-p) > 5

In our question, n = sample of customers = 303

p = probability that customers will take free sample = 56%

So, np = $303 \times 0.56$ = 169.68 > 5

n(1-p) = $303 \times (1-0.56)$ = 133.32 > 5

Since, both conditions are satisfied so it is appropriate to use the normal approximation to the binomial.

Now, mean of the normal distribution is given by;

Mean, $\mu$ = $n \times p$ = 169.68

Also, the standard deviation of the normal distribution is given by;

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.56 \times (1-0.56)}$ = 8.64

Let X = Number of people who will take your free sample

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) The probability that more than 180 will take your free sample is given by = P(X > 180) = P(X > 180.5) {Using continuity correction}

P(X > 180.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{180.5-169.68}{8.64}$ ) = P(Z > 1.25) = 1 - P(Z < 1.25)

= 1 - 0.8944 = 0.1056

(b) The probability that fewer than 200 will take your free sample is given by = P(X < 200) = P(X < 199.5) {Using continuity correction}

P(X < 199.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{199.5-169.68}{8.64}$ ) = P(Z < 3.45) = 0.9997

(c) We are given in the question that of those who take the free samples, about 39% will buy what they have sampled, this means that we have;

P(Buy the product / taken a free sample) = 0.39

So, Probability(customer will take a free sample and buy the product) = P(customer take a free sample) $\times$ P(Buy the product / taken a free sample)

= 0.56 $\times$ 0.39 = 0.2184

(d) Now our mean and standard deviation will get changed because the probability of success now is p = 0.2184 but n is same as 303.

So, Mean, $\mu$ = $n \times p$ = $303 \times 0.2184$ = 66.18

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.2184 \times (1-0.2184)}$ = 7.192

Now, the probability that between 60 and 80 customers will take the free sample and buy the product is given by = P(60 < X < 80) = P(59.5 < X < 80.5) {Using continuity correction}

P(59.5 < X < 80.5) = P(X < 80.5) - P(X $\leq$ 59.5)

P(X < 80.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{80.5-66.18}{7.192}$ ) = P(Z < 1.99) = 0.9767

P(X $\leq$ 59.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{59.5-66.18}{7.192}$ ) = P(Z $\leq$ -0.93) = 1 - P(Z < 0.93)

= 1 - 0.8238 = 0.1762

Therefore, P(59.5 < X < 80.5) = 0.9767- 0.1762 = 0.8005.

4 0

3 years ago

It took 7 hours to mow four equal-size lawns. At that rate, how many lawns could be mowed in 35 hours?

NISA [10]

Answer: 20 lawns could be mowed in 35 hours.

Step-by-step explanation: Lets write a ratio to represent this problem. The ratio 7:4 states that for every 7 hours you will mow 4 lawns.

Now we have to figure out how many lawns we will mow in 35 hours. Simply keep adding 7 to the left side of the ratio and 4 to the right side.

7:4

14:8

21:12

28:16

35:20

So in 35 hours, 20 lawns will be mowed.

7 0

3 years ago

help asap pls!!!!!!!!!! Which equation can be used to determine the distance between the origin and (–2, –4)?

Semenov [28]

We know that
the distance formula is
d=√[(y2-y1)²+(x2-x1)²]
point (0,0) and point (–2, –4)
d=√[(-4-0)²+(-2-0)²]---------> d=√[16+4]-------> d=√20
d=4.47 units

the answer is the option B
see the attached figure

8 0

3 years ago