Ask question

Ask question

All categories

2 years ago

15

Find the probability that the next student will select hamburger for lunch. You

1 answer:

adell [148]2 years ago

6 0

Answer=3/10

Step wise =

Total number of student=40

student who eat Hamburger=12

now,

Probability of eating hamburger student = 12/40

=3/10

You might be interested in

Determining function values from a graph

professor190 [17]

Answer:

The y value of a point where a vertical line intersects a graph represents an output for that input x value. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output.

6 0

3 years ago

Please help me with this problem

xxTIMURxx [149]

Answer:

1516

Step-by-step explanation:

hope can help you

2000x 2%=40

9000x 5%= 450

(30000-9000-2000) x 5.4% = 1026

total tas 40+450+1026=

8 0

1 year ago

46 cents is what fraction of a dollar

Vikentia [17]

Since there are a 100 cents in a dollar. the 100 is the denominator.
46/100, reduce; 23/50, can't be reduced any further. so 23/50 is the fraction.

7 0

2 years ago

Read 2 more answers

Jaden is building a fence around his yard. He wants to know how much fencing he needs. His yard is shaped like a rectangle that

Ivahew [28]

Answer:258

Step-by-step explanation:

4 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago