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3 years ago

Find the unit rate (How much for

Mathematics

1 answer:

Ber [7]3 years ago

8 0

Answer:

$16.455

Step-by-step explanation:

98.73/6 = y

16.455 = y

You might be interested in

Divide £70 in the ratio 4:3

kirill115 [55]

Answer:

Total=4 +3=7

For the 4

4÷7×70=£40

For the 3

3÷7×70=£30

So the division of £70 in ratio 4:3 is £40:£30

7 0

3 years ago

Which is the angle measure represented by 7.5 rotations clockwise?

kondaur [170]

Recall that 1 rotation is the equivalent of 360 degrees or 2pi radians.

7.5 rotations clockwise 2pi rad
------------------------------- * ---------------------------------------
1 1 rotation counterclockwise

= -47.12 radians, or -2700 degrees.

8 0

4 years ago

Jessica had $150 in her savings account after her first week of work. She then started adding $35 each week to her account for t

Fofino [41]

35w + 150 = s

$185, $220, $255, $290, $325

(Add $35 each week for 5 weeks)

6 0

3 years ago

Two numbers are in the ratio 7:5 and their difference is 22.Find the numbers

omeli [17]

Answer:

Let x and y be the two numbers. Then

yx=57 and x−y=20

Cross multiplying the proportion

5x=7y or y=75x

So, x−y=20

x−75x=20⇒77x−5x=72x=20

⇒x=2140=70

y=75x=75×70=50

So smaller number is 50.

Check :yx=5070=57 and 70−50=20

4 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago