How many milliliters are there in 0.5 liters?
5.000
50
500
5
Correct Answer : C
We use the relation 1 L=1000 mL to convert 0.5 L to mL as follows.
Thus, 0.5 L is 500 mL.
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Related Questions
Correct Answer is C
Explanation
The question requires us to find the percentage change in weight of a person.
First, we need to find the change in weight over the 3 months
Change in weight= 180-160=20 lb
Percent change in weight is change of original weight *100. Thus
The percent change in weight is 11% to the nearest whole number.
Correct Answer is B
Explanation
We need to find the circumference of the circle from the given area.
To find the circumference, we need to find the radius of the circle from the given area.
Let r be the radius of the circle, and the area of the circle is given by
Substituting the value of area in the above equation becomes
Dividing both sides by pi and taking square root on both sides yields
The radius of the circle is 7 in. and the circumference of the circle is given by the relation:
Â
Correct Answer is C
Explanation
The number of boxes is determined by finding the volume of the room divided by the volume of the box.
Number of boxes
The approximate number of boxes that can be stored in the room is 92.
Correct Answer is C
Explanation
We need to find the price changes in each day of the week in order to find the price of the stock on Thursday.
Monday’s price was $50.33
Tuesday’s price went up by $2.35 from Monday’s price. The price was $(50.33+2.35)=$52.68
Wednesday’s price decreased by $1.07 from Tuesday’s price, and so the price of the stock was $(52.68-1.07)=$51.61
Thursday’s price increased by $0.75 from Wednesday’s price, which was $(51.61+0.75)=$52.36
Therefore, the price of the stock on Thursday was $52.36.
Correct Answer is C
Explanation
We required to find the units of finding the weight of a spoon. The mass is measured in kilogram and grams. Kilograms are used to measure the mass of heavier objects while grams for small and lighter objects.
Thus, gram is the appropriate unit for measuring weight of a spoon.
Correct Answer is A
Explanation
To find the greatest number from the given options, we first convert the decimal numbers into fractions.
-0.7 becomes 7/10
-1.3 becomes 13/10
Then, we find the LCM for the denominators of the given fractions. The LCM of 3 and 10 is 30. Now we can multiply each fraction with the LCM.
-2/3*30=-20
-7/10*30=-70
-13/10*30=-39
-4/3*30=-40
Comparing the obtained values from above, -20 is the greatest followed by -39, -40, -70 in that order. The fraction -2/3 gave a value of -20, which was the greatest value. Thus, -2/3 is the greatest value from the given option.
Correct Answer is A
Explanation
We follow the order of operations to solve the given expression.
First, we start with the numerator and solve it as follows
[2(3+5*3)]
We start with multiplication in inner brackets, 5*3=15. The expression becomes
[2(3+15)]
Then, we conduct the addition of 3+15=18. Then, the expression yields
[2(18)]=2*18=36
Now, we solve for denominator, which is 12/2=6.
Thus, the expression is reduced into
The expression reduces into 6.
Correct Answer is B
Explanation
We need to find the value of x from the given equation. First, we move the value of 10 to the right-hand side of the equation.
Add 10 to both sides of the equation
Next, we apply the absolute rule:
,  a>0, then u=a or u=-a
In this case a=12, which is greater 0. Then, the first condition becomes
Solving for x
The second condition becomes
Solving for x
Then, the value of x is -1 or 2.
Correct Answer is B
Explanation
We use the calculator to determine the positive square root of 20, which is then multiplied by 20.
Using the calculator,
Multiplying the square root above with 10 becomes
The approximate value is 44.7.
Correct Answer is D
Explanation
We let x represent the amount of vanilla in mL, since this is what the question is asking us to find.
Next, we will set up a proportion with number of teaspoons on the numerator and amount in mL in the denominator.
Cross-multiply to find the value of x
A recipe of 2.5 teaspoons equals 12.325 mL.
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