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Andrej [43]
1 year ago
6

I am in need of some help :(It looks so simple yet I don't get how to do it. Can anyone help explain?(picture of question attach

ed)source: Pearson/Savvas Realize

Mathematics
1 answer:
Firdavs [7]1 year ago
5 0

x = 14

Explanation:

height = x

base 1 = 98

base 2 = 2

Using altitude formula in geometric mean:

base 1/height = height/base 2

98/x = x/2

\begin{gathered} \frac{98}{x}=\frac{x}{2} \\ \text{cross multiply:} \\ 2(98)\text{ = x}\times x \end{gathered}\begin{gathered} 196=x^2 \\ \text{square root both sides:} \\ \sqrt[]{196}\text{ = }\sqrt[]{x^2} \\ x\text{ = }\sqrt[]{196} \\ x\text{ = 14} \end{gathered}

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I would really like to have some help and know how to do this
NISA [10]
If K=C+273.15
and in a. C=200° then K=200+273.15
8 0
3 years ago
Text wants to make a lawn in front of his house. He buys enough sod to make a 10 ft.by 30 ft. rectangle section of sod for $200
Trava [24]
This isn't a question. What is the question?
3 0
3 years ago
THIS IS URGENT I AM BEING TIMED IF YOU ANSWER ALL OF THEM I WILL MARK YOU BRAINLYIST.
Dafna1 [17]

Answer:

The answer is 6.56

Step-by-step explanation:

11.20 is the total.  You multiply that by .15 and get $1.68 for the tip.  You multiply $11.20 by .05 for the tax and get  $.56.  You  add them together and get $13.44.  You subtract that from $20.00 and get $6.56

7 0
3 years ago
Jessica is getting ready for college she needs to have an overall grade of 85 in her Math class to qualify for the college that
VLD [36.1K]
So,

Her average so far is the sum of the five scores divided by the number of tests she has taken so far.

(87 + 81 + 85 + 90 + 78)/5 = 84.2

(87 + 81 + 85 + 90 + 78)/5 < 85 :(

She will not get into that college if she doesn't increase her average on the last test.

85 - 84.2 = 0.8

0.8(5) = 4.0

On her last test, she needs to score 4 points higher than her current average in order to get into the college.

84.2 + 4 = 88.2

Jessica needs to score at least 88.2 on her last test.

5 0
3 years ago
In a sample of 1200 U.S.​ adults, 191 dine out at a resaurant more than once per week. Two U.S. adults are selected at random
allochka39001 [22]

Answer:

a) The probability that both adults dine out more than once per week = 0.0253

b) The probability that neither adult dines out more than once per week = 0.7069

c) The probability that at least one of the two adults dines out more than once per week = 0.2931

d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Step-by-step explanation:

In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.

Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592

Now, assuming this probability per person is independent of each other.

Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.

a) The probability that both adults dine out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult A and adult B dine out more than once per week = P(A n B)

= P(A) × P(B) (since the probability for each person is independent of the other person)

= 0.1592 × 0.1592

= 0.02534464 = 0.0253 to 4 d.p.

b) The probability that neither adult dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

Probability that neither adult dines out more than once per week = P(A' n B')

= P(A') × P(B')

= 0.8408 × 0.8408

= 0.70694464 = 0.7069 to 4 d.p.

c) The probability that at least one of the two adults dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

The probability that at least one of the two adults dines out more than once per week

= P(A n B') + P(A' n B) + P(A n B)

= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]

= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)

= 0.13385536 + 0.13385536 + 0.02534464

= 0.29305536 = 0.2931 to 4 d.p.

d) Which of the events can be considered unusual? Explain.

The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).

And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Hope this Helps!!!

6 0
3 years ago
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