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Kamila
11 days ago
10

Nine and one-half less than four and one-half times a number is greater than 62.5. Which of the following represents the solutio

n set of this problem?
(16, positive infinity)
(Negative 16, positive infinity)
(Negative infinity, 16)
(Negative infinity, Negative 16)
Mathematics
2 answers:
lawyer [9.2K]11 days ago
8 0
16, + infinity. Multiply both sides by 1/9 n extends to infinity from 16.
tester [8.8K]11 days ago
4 0

Answer:

{16, +infinity}

Step-by-step explanation:

Represent the number with n

Required

Find the solution set

According to the question, it states that;

4½n - 9½ > 62.5

Transform all numbers into decimals

4.5n - 9.5 > 62.5

Add 9.5 to both sides

4.5n - 9.5 + 9.5 > 62.5 + 9.5

4.5n > 72

Multiply both sides by 2

2(4.5n) > 72 * 2

9n > 144

Multiply both sides by ⅑

⅑ * 9n > 144 * ⅑

n > 16

This means that n ranges from positive 16 to infinity;

Hence, the solution set is {16, +infinity}

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Explain why the following expression is false. |x| < -4
Zina [9157]

Step-by-step explanation:

When a negative number is placed within a modulus function, the result will be positive. For instance, |-3| equals 3, |-6| equals 6, and |5| equals 5, etc.

A modulus function, expressed as |x|, is always positive unless x is zero, in which case it equals zero.

Consequently, |x| cannot be less than -4 because |x| is always non-negative. Thus, the statement is inaccurate.

8 0
20 days ago
Yoku is putting on sunscreen. He uses 2\text{ ml}2 ml2, start text, space, m, l, end text to cover 50\text{ cm}^250 cm 2 50, sta
AnnZ [9071]

Answer:

13 Millilitres.

Step-by-step explanation:

Yoku applies 2ml to cover 50cm^2 of his skin.

He seeks to find out how many millilitres he needs to cover 325cm^2 of his body.

With 2ml, Yoku covers 50cm^2 of his skin

He will use \frac{2}{50}ml of sunscreen for covering 1cm^2 of his skin.

Thus:

For covering 325cm^2 of his body, he will require:

\frac{2}{50} X 325 ml\\=13ml

he would need 13 Millilitres.

7 0
27 days ago
Read 2 more answers
if a student scored 78 points on a test where the mean score was 84.5 and the standard deviation was 3.1. the student's z score
Inessa [8989]

Answer:

(value - mean)/standard deviation.

Now substitute 78 for the value, 84.5 for the mean, and 3.1 for the standard deviation.

Step-by-step explanation:

5 0
25 days ago
1) Jodi liked to collect stamps. On 3 different days she bought 6 stomps. Then she
babunello [8402]

Answer:

4

Step-by-step explanation:

6-4+2*5=12

12/3

4

6 0
6 days ago
At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
Zina [9157]

Answer:

(a1) The chance that the temperature rises by less than 20°C is 0.667.

(a2) The probability of the temperature increase falling between 20°C and 22°C is 0.133.

(b) The likelihood that the temperature increase could be hazardous at any moment is 0.467.

(c) The anticipated value of the temperature rise is 17.5°C.

Step-by-step explanation:

Let X denote the temperature increase.

The random variable X is distributed uniformly over the interval [10°C, 25°C].

The probability density function for X is shown here:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

The probability of the temperature increase being under 20°C can be calculated as follows:

P(X

Consequently, the chance that the temperature increase will be below 20°C is 0.667.

(a2)

The probability of the temperature rise being in the range from 20°C to 22°C is computed as follows:

P(20

This leads to the probability of the temperature increase being between 20°C and 22°C being 0.133.

(b)

To find the probability that the increase in temperature could be dangerous, we calculate:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

This results in a probability of 0.467 that the temperature rise is potentially dangerous at any time.

(c)

The expected value of the uniform random variable X is determined as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

The expected value for the temperature increase computes to 17.5°C.

7 0
23 days ago
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