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musickatia [10]
2 years ago
5

7. The scale on a map of Florida is 1 ⅛ inches to 25 miles. The distance on the map from Gainesville to Ocala is 1 ⅝ inches. Bas

ed on this information, approximately how many miles are there between the two cities?
A=14
B=17
C=36
D=41
Mathematics
1 answer:
sesenic [268]2 years ago
8 0
C is the answer I think
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Are these lines perpendicular? yes or no?<br> y= 4x+2<br> y= (-1/4)x+12
antoniya [11.8K]

Answer: yes

Step-by-step explanation:

3 0
3 years ago
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Please help with #35 I don’t get it and I’m really stressed
skad [1K]

Don't let it get to ya. If they didn't think you're capable of answering this question, they wouldn't ask you. So they know your abilities better than you do.

We don't know the unknown number, so in order to discuss it, we have to give it some kind of a name. We could call it 'x', 'y', 'Ana, 'Q', or 'Ralph'. Let's call it 't'.

The unknown number . . . . . t

Five times the number . . . . . 5t

Nine less than five times the number . . . . . 5t - 9 . Hold on to this quantity.

Twice the number . . . . . 2t .

Three more than twice the number . . . . . 2t + 3. Hold on to this quantity too.

The question says that these two quantities are equal. So the equation is

5t - 9 = 2t + 3 . This is the equation in both choices 'c' and 'd' . I guess we'll have to solve it to find out which choice is the correct one.

The equation is . . . . . 5t - 9 = 2t + 3

Add 9 to each side . . . . . 5t = 2t + 12

Subtract 2t from each side . . . 3t = 12

Divide each side by 3 . . . . . t = 4 .

It looks like choice-d is the winner.

3 0
3 years ago
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The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 900lbs.
katen-ka-za [31]

Answer:

0.0369

Step-by-step explanation:

normalcdf (1220,1320,900,200) is 0.0369

3 0
2 years ago
Weights were recorded for all nurses at a particular hospital, the mean weight for an individual nurse was 135 lbs. with a stand
Aloiza [94]
Given:
population mean, &mu; =135
population standard deviation, &sigma; = 15
sample size, n = 19

Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.

Estimate of sample mean
\bar{x}=\mu=135

Estimate of sample standard deviation
\s=\sqrt{\frac{\sigma^2}{n}}
=\sqrt{\frac{15^2}{19}}=3.44124  to 5 decimal places.

Thus, using the normal probability table,
P(125
=P(\frac{125-135}{3.44124}
=P(-2.90593
=P(Z
=P(Z

Therefore 
The probability that the mean weight is between 125 and 130 lbs 
P(125<X<130)=0.0731166-0.0018308
=0.0712858



3 0
2 years ago
What is the equation of the line that passes through the points (2, –1) and (6, 1)?
Zolol [24]
Y=kx+b
2k+b=-1
6k+b=1
6k+b-2k-b=2
4k=2
k=1/2
b=-2
y=1/2x-2
So your answer is option A

8 0
3 years ago
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