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4 years ago

15

Lee class has 1.4 times as many students as Fred's. If there are 60 students, how many students does Fred's class have? How many

students does Lee's class have?

1 answer:

slava [35]4 years ago

8 0

I think Lee has 42
Fred has 30

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Follow the steps to find the value of x 1/5(x)5/1=6/3(5/1)

german

Answer: x=10

This is the answer I got.

6 0

3 years ago

The high temperature in Fairbanks, Alaska was 15.7 °F one day. That night, the temperature fell 38.4 degrees. What was the low t

enot [183]

Answer:

-22.7

Step-by-step explanation:

Since the temperature fell, that would means that it's was cooler outside.

Hotter = Temperature goes up

Colder = Temperature goes down

15.7 - 38.4 = -22.7

3 0

3 years ago

Can someone do this please?

amid [387]

Answer:

<1 =73

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

72+ 35 + <1 = 180

Add like terms

107 + <1 = 180

Subtract 107 from each side

<1 = 180-107

<1 =73

8 0

2 years ago

The area of a triangular block is 49 square inches. If the base of the triangle is twice the height, how long are the base and t

andrezito [222]

1) Formula: area = height * base / 2

2) Call x the height:

height = x

base = 2x

3) State the equation:

(x)(2x) = 49 in^2

4) Solve the equation:

2x^2 = 49 in^s

x^2 = (49/2) in^2

x = √(49/2) in

x = 7√2 / 2 in = 3.5 √2 in

5) Solution:

height = 3.5 √2 in and base = 7√2 in

6) Verification: area = (7√2 in) (3.5√2 in) = 49 in^2

Answer: height = 3.5√2 in and base = 7√2 in.

5 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago