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4 years ago

What is 55/100 simplified

Mathematics

2 answers:

Ivanshal [37]4 years ago

6 0

11/20 because 5 multip by 11=55and 5 multiple by 30=50

Luden [163]4 years ago

3 0

55/100 can be simplified to 11/20

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A movie theater sells only two types of tickets: adult tickets for $20 each and student tickets for $12 each. One weekend, the t

Yakvenalex [24]

Answer:

adults tickets=1384

kids:616 tickets

1384-616= 768 tickets

Step-by-step explanation:

let x be adult tickets and y kids

x+y=2000 ⇒x=2000-y

20x+12y=35072 solve by substitute for x=2000-y

20(2000-y)+12y=35072

40000-20y+12y=35072

-8y=35072-40000

-8y=-4928

y=-4928/-8

y=616

x=2000-616

x=1384 tickets

check:

20(1384)+12(616)=35072 ( correct)

4 0

4 years ago

In triangle XYZ, mZZ > mZX + mZY. Which must be true about AXYZ?

Semmy [17]

Answer:

D. m∠X + m∠Y < 90°

Step-by-step explanation:

<h3>Given</h3>

ΔXYZ with m∠Z > m∠X + m∠Y

<h3>Options</h3>

incorrect, the sum of these two angles should be > 90°

incorrect, it should be < 90°

<u>∠X and ∠Y are complementary</u>

incorrect as angle Z is greater that the sum of angles X and Y, then sum of three angles would be greater than 180°

correct, this is the only possible option

4 0

3 years ago

clark made a model of his house. his house is 30 1/2 feet long. the dimensions of the model were 1/25 the dimensions of clarks a

balandron [24]

Answer:

1.22 feet, or 1 11/50 ft

Step-by-step explanation:

Find 1/25 th of 30 1/2 feet:

1 61 ft

----- * ---------- = 1.22 feet, or 1 11/50 ft

25 2

The length of the model is 1.22 feet, or 1 11/50 ft

7 0

4 years ago

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The

adelina 88 [10]

Answer:

The 90% confidence level is $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 64$

The mean age is $\= x = 20 \ years$

The standard deviation is $\sigma = 4 \ years$

Generally the degree of freedom for this data set is mathematically represented as

$df = n - 1$

substituting values

$df = 64 - 1$

$df = 63$

Given that the level of confidence is 90% the significance level is mathematically evaluated as

$\alpha = 100 - 90$

$\alpha =$ 10 %

$\alpha = 0.10$

Now $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

$t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

The margin for error is mathematically represented as

$MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

$MOE = 1.699 * \frac{4 }{\sqrt{64} }$

$MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

$\= x - MOE < L < \= x + E$

substituting values

$20 - 0. 85 < L < 20 + 0.85$

$19.15< L < 20.85$

4 0

3 years ago

NEED HELP ASAP

lora16 [44]

Answer:

the planes ground speed is 658.94 kph

Step-by-step explanation:

4 0

3 years ago

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