Carlon plans to evaporate a number of gallons of water from 90 gallons of a 15% salt solution to make a 25% salt solution. which

To solve this problem, simply sum all of the grocery expenses and then divide that total by 4...

∠ROT=160°

∠SOT=100°

Now we calculate ∠SOR as follows: ∠SOR = ∠ROT - ∠SOT

∠SOR = 160° - 100°

∠SOR = 60°

It is stated that the angles ROQ, QOS, and POT all have the same measure.

Thus, ∠SOQ + ∠QOR = 60°

Since ∠SOQ equals ∠QOR, we can express this as:

2∠SOQ = 60°

From which we find ∠SOQ = 60° ÷ 2

∠SOQ = 30°

Also, ∠POT = ∠SOQ = ∠ROQ = 30°

Let x represent the number of caps

cost per cap = $6

cost for x caps = 6x

shipping charge = $25

overall budget = $1000

We can set up the following inequality:

Because the total amount cannot exceed 1000, we have

25 + 6x ≤ 1000

6x ≤ 975

x ≤ 162.5

rounding,

x ≤ 163

This means she can purchase a maximum of 163 caps.

Answer:

Step-by-step explanation:

Hello!

The monitoring system alerts the driver when the vehicle's tire pressure drops to 28% below the set target pressure.

Let X be: the target tire pressure for a particular vehicle (measured in pounds per square inch)

a)

X= 28 psi

When the monitoring device alerts at a pressure of 28% below the designated target: X-0.28X

Initially, calculate 28% of 28 psi.

28*0.28= 7.84

Next, subtract this 28% calculation from the target pressure:

28 - 7.84= 20.16

The TPMS will activate its warning at 20.16 psi.

b)

Assuming X~N(μ;σ²)

μ= 28 psi (as the average indicates accurate targeting, this represents the expected tire pressure)

σ= 3 psi

P(X≤20.16)

The standard normal distribution is available in tables. To convert any random variable X with a normal distribution, one subtracts its mean and divides by the standard deviation.

To compute the desired probabilities, the variable value undergoes transformation to fit the standard normal distribution Z, after which standard normal tables are referenced to find probabilities.

Z= (X-μ)/σ= (20.16-28)/3= -2.61

Now locate the probability corresponding to the Z value using the Z-table. Given the negative result, refer to the left entry; in the first column, find the integer and first decimal for -2.6- and in the first row locate the second decimal for -.-1

The probability link for -2.61 is:

P(Z≤-2.61)= 0.005

c)

The task is to determine the likelihood of encountering a tire at random within the recommended inflation range, expressed as:

P(30≤X≤26)= P(X≤30)-P(X≤26)

Determine both Z values:

Z= (30-28)/3= 0.67

Z= (26-28)/3= -0.67

P(Z≤0.67)-P(Z≤-0.67)= 0.749 - 0.251= 0.498

I hope this information is useful!