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3 years ago

Find the difference of 18.6 and 5.24.*

Mathematics

1 answer:

Charra [1.4K]3 years ago

5 0

Answer:

13.36

Step-by-step explanation:

18.60 - 5.24

=>13.36

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2. Trouvaughn bought b bags of oranges, which cost 80 cents per bag. If Trouvaughn paid with a

Damm [24]

Amount that trouvaughn paid : $0.80b
ten dollar bill = $10

so $(10-0.80b)

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3 years ago

In the figure below, angle y and angle x form vertical angles. Angle y forms a straight line with the 60° angle and the 70° angl

Lynna [10]

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4 0

3 years ago

Suppose j varies jointly with g and v, and j=2 when g=4 and v=3. Find j when g=8 and v=9.

vladimir2022 [97]

Answer:

$j=12$

Step-by-step explanation:

we know that

In a join variation, If j varies jointly with respect to g and v, the equation will be of the form $j = kgv$

where k is a constant

step 1

Find the value of k

we have

j=2,g=4,v=3

substitute and solve for k

$2 = k(4)(3)$

$2 = 12k$

$k=\frac{1}{6}$

The equation is equal to

$j=\frac{1}{6}gv$

step 2

Find the value of j when g=8,v=9

substitute the values in the equation and solve for j

$j=\frac{1}{6}(8)(9)$

$j=12$

3 0

3 years ago

Evaluate -a - 12a^2 for a = -1 a. -13 b. 13 c.-11

Gelneren [198K]

substitute a = - 1 into the expression

- (- 1) - 12(- 1)² = 1 - (12 × 1 ) = 1 - 12 = - 11 → C

8 0

3 years ago

Read 2 more answers

12% of children are nearsighted, but this condition often is not detected until they go to kindergarten. A school district tests

Klio2033 [76]

Answer:

There is a 34.60% probability that 0 or 1 of them is nearsighted.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they are nearsighted, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

12% of children are nearsighted. This means that $p = 0.12$ .

A school district tests all incoming kindergarteners' vision. In a class of 18 kindergarten students, what is the probability that 0 or 1 of them is nearsighted?

There are 18 students, so $n = 18$