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Mekhanik [1.2K]
2 years ago
5

SOMEONE PLEASE ANSWER I NEED THE ANSWER URGENTLY!!!! PLEASE ILL GIVE U BRAINLIEST I JUST NEED ANSWERS!!!

Mathematics
1 answer:
Dafna1 [17]2 years ago
5 0
I can’t see anything with the pictures
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What is the length of the hypotenuse in a right triangle with legs of length 7 and 24?
Fiesta28 [93]
A² + b² = c²   a and b are the legs and c is the hypotenuse.

7² + 24² = c²

49 + 576 = c²

625 = c²

√625 = √c²

25 = c

hope that helps, God bless!

4 0
2 years ago
+<br> What is the answer
padilas [110]
The answer is edition
8 0
3 years ago
I’m not sure why the picture didn’t work
lutik1710 [3]

Answer:

4.1

Step-by-step explanation:

6 0
2 years ago
Which is the best estimate to the nearest one percent of startfraction 7 over 15 endfraction?
ipn [44]

The best estimate to the nearest one percent of the fraction; 7/15 is; 47%.

<h3>What is the best estimate of the fraction to the nearest percent?</h3>

From the task content, it follows that the fraction given whose estimate is to be determined is; 7/15.

The fraction expressed as a percentage is;

(7/15) × 100 %

= 46.667%.

Hence, when rounded to the nearest one percent; = 47%.

Read more on percentage;

brainly.com/question/843074

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4 0
1 year ago
Merta reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time.
kenny6666 [7]

Answer:

No, the on-time rate of 74% is not correct.

Solution:

As per the question:

Sample size, n = 60

The proportion of the population, P' = 74% = 0.74

q' = 1 - 0.74 = 0.26

We need to find the probability that out of 60 trains, 38 or lesser trains arrive on time.

Now,

The proportion of the given sample, p = \frac{38}{60} = 0.634

Therefore, the probability is given by:

P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]

P(p\leq 0.634) = P[z\leq -1.87188]

P(p\leq 0.634) = P[z\leq -1.87] = 0.0298

Therefore, Probability of the 38 or lesser trains out of 60 trains to be on time is 0.0298 or 2.98 %

Thus the on-time rate of 74% is incorrect.

6 0
2 years ago
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