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3 years ago

Subtracting decimals 4 - 21

Mathematics

1 answer:

maw [93]3 years ago

4 0

Answer:

-17

Step-by-step explanation:

It is basically subtracting 4 from 21 except the answer is negative

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Find the slope and the y-intercept of the graph of y=11+1.5x.

Anni [7]

Honestly man i don’t know Im in 6th grade bro

6 0

3 years ago

Read 2 more answers

Please help me with this!! It might be very easy for you I will also give brainliest if correct

Nimfa-mama [501]

The answer is always

6 0

3 years ago

Soit trois nombres consécutifs. Lorsque l’on soustrait le double du deuxième nombre du quadruple du

irina [24]

Answer:

Les trois nombres sont 5, 6 et 7.

Step-by-step explanation:

Soit x le premier nombre.

x + 1 sera le second et x + 2 sera le troisième.

On pose l'équation :

2x + 3(x + 1) = 4(x + 2)

2x + 3x + 3 = 4x + 8

5x - 4x = 8 - 3

x = 5

4 0

3 years ago

Solve the following equations x1/3=2 <br> Y-2=9<br> (2s)1/2=9<br> 2n-1=16

Nookie1986 [14]

$x \times \frac{1}{3} = 2 \\ = > x = 2 \times 3 \\ = > x = 6$

$y - 2 = 9 \\ = > y = 9 + 2 = 11$

$(2s) \frac{1}{2} = 9 \\ = > s = 9$

$2n - 1 = 16 \\ = > 2n = 16 + 1 \\ = > 2n = 17 \\ = > n = \frac{17}{2} = 8.5$

<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />

7 0

2 years ago

Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

$P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}$

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

$P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483$

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

5 0

3 years ago