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torisob [31]
3 years ago
13

Work out the missing area in each pair of similar shapes (Q.2)

Mathematics
1 answer:
Temka [501]3 years ago
4 0
270 ÷ 3=90 so the answer is 90cm^2
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6÷0.9 help plzzzzzzzzzzzzzzzzzzz
Rus_ich [418]

6 / .9 *10/10

60/9

9 goes into 60 6 times with 6 left over

6 6/9

6 2/3

8 0
4 years ago
Read 2 more answers
Evaluate the expression 5t + 2m = 7 and t =2 HELPP
Tomtit [17]

To solve, you need to plug in the values they give you into the equation.

5t + 2m

Plug in the numbers they give you

5(2) + 2(7)

Solve.

10 + 14

Simplify.

24

6 0
3 years ago
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HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Mnenie [13.5K]

Answer:

The correct answer is D

Step-by-step explanation:

5 0
4 years ago
Which values of c will cause the quadratic equation –x2 3x c = 0 to have no real number solutions? check all that apply.
nika2105 [10]

The given quadratic equation will not have any real solution for c<-9/4.

The given quadratic equation is:

-x^{2} +3x+c=0

<h3>What is a quadratic equation?</h3>

Any equation of the form ax^{2} +bx+c=0 is called a quadratic equation with a≠0.

In order to have no real solution, the discriminant of a quadratic equation will be less than zero.

D < 0

3^{2} -4(-1)(c) < 0

9+4c < 0

c < -\frac{9}{4}

For c < -\frac{9}{4} the given quadratic equation will have no real solutions.

Hence, the given quadratic equation will not have any real solution for c<-9/4.

To get more about quadratic equations visit:

brainly.com/question/1214333

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
4 years ago
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