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2 years ago

12 is 75% of what number

1 answer:

3 0

This is a really easy problem.

All you have to do is do 12 divided by 75%.

If you don't have a calculator, turn 75% into 3/4
3/4 times 12 which is

16
I hope i helped!:)

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Anyone knows how to solve -6(-5x-3)-5x+17?

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2 years ago

Solve: logx64 = 3 which is an equivalent equation?

zepelin [54]

By applying the definition of logarithm, you have

$\log_x(64) = 3 \iff x^3 = 64$

So, if you consider the cubic root of both sides, you have

$x = \sqrt[3]{64} = 4$

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2 years ago

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Select and use the most direct method to solve 2x(x + 1.5) = -1. Describe and justify the methods you used to solve the quadrati

miskamm [114]

Step-by-step explanation:

Given that,

A quadratic equation,

2x(x + 1.5) = -1

We need to solve the quadratic equation. Firstly we need to simplify the above equation to form it as $ax^2+bx+c=0$ .

So,

$2x^2+3x=-1\\\\2x^2+3x+1=0$

Here, a = 2, b = 3 and c = 1

The roots of the given equation can be given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Putting all the values we get :

$x=\dfrac{-b\pm \sqrt{b^2+4ac} }{2a}, \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-3\pm \sqrt{3^2+4(2)(1)} }{2(2)}, \dfrac{-3\pm \sqrt{3^2-4(2)(1)} }{2(2)}\\\\x=\dfrac{-3+1}{4}, \dfrac{-3-1}{4}\\\\x=-\dfrac{1}{2}, -1$

So, the roots of the given equation is -1/2 and -1.

5 0

2 years ago

Need help pleaseee x

Grace [21]

$-95/5=-19$

$70/-2=-35$

$-18 \times 10=-180$

$27 \times -3=-81$

7 0

2 years ago

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The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

2 years ago