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Mademuasel [1]
2 years ago
9

2.14x - 42.9 for x = 22.4 ? A. -18.36 B. 5.026 C. 47.939 D. 90.839

Mathematics
1 answer:
Fynjy0 [20]2 years ago
7 0

Answer:

Step-by-step explanation:

Your answer would be B., because you would multiply

2.14 x  22.4   =

47.939  

then you will subtract that by 42.9

47.939 - 42.9 =  

5.026

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What number has 2 more tens than 40 and the same number of ones as 18
Juli2301 [7.4K]
The answer is 68.
To solve 2 more tens than 40, you add 20 to 40, which is 60.
To solve for the same number of ones as 18, then it would only be 68.

Hope this helps!!
5 0
2 years ago
Which series of transformations will not map figure H onto itself?
chubhunter [2.5K]

Answer:

(x + 0, y − 2), reflection over y = 1

Step-by-step explanation:

H will be shifted down 2 units and then reflected over y = 1, which maps H into itself

3 0
2 years ago
R is inversely proportional to A R = 12 when A = 1.5 a) Work out the value of R when A = 5 b) Work out the value of A when R = 9
lapo4ka [179]

Answer:

a) R = 3.6

b) A = 2

Step-by-step explanation:

To find the value, start by modeling the inverse proportionality by using the base equations with the original givens.

y = k/x

12 = k/1.5

18 = k

Now we use that to model the equation

R = 18/A

And we can now use that to solve parts a) and b)

a)

R = 18/A

R = 18/5

R = 3.6

b)

R = 18/A

9 = 18/A

2 = A

5 0
2 years ago
) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
2 years ago
Andrew bashir and candy are trying to save money for a birthday party.If andrew saves 1/4 of the total needed bashir saves 2/5 a
DerKrebs [107]

Answer:

1/4.

Step-by-step explanation:

That is 1 - (1/4 + 2/5 + 1/10).

The Lowest common multiple  4 5 and 10 is 20 so we have

20/20 - (5/20 + 8/20 + 2/20)

=  20/20 - 15/20

= 5/20

= 1/4 answer.

5 0
2 years ago
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