All categories

2 years ago

Jenny, Keith, and Joanna participated in a mini-

Mathematics

1 answer:

Elan Coil [88]2 years ago

4 0

Answer:

Jenny 1.75 miles, Keith 4.75 miles, Jonna 9.5 miles

Step-by-step explanation:

Distance per person:

Jenny - x, Keith - y, Joanna - z

Equation as per given:

y = x + 3
z = 2y = 2(x + 3)
x + y + z = 16

Substitute and solve for x:

x + (x + 3) + 2(x + 3) = 16
2x + 3 + 2x + 6 = 16
4x = 7
x = 7/4 = 1.75

Then

y = 1.75 + 3 = 4.75
z = 4.75*2 = 9.5

You might be interested in

The figure below is a parellelogram. What type(s) of symmetry does it have?

Vinvika [58]

Line symmetry I think

5 0

2 years ago

2x+2y= 24 x=3y What is the value of x+y?

natita [175]

Answer:

x = 9

y = 3

Step-by-step explanation:

2x + 2y = 24 is the same as x + y = 12

using substitution, x = 12-y

12 - y = 3y

12 = 4y

y = 3

x = 3(3)

x = 9

5 0

2 years ago

Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa

Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : $\mu = 15.45$

Standard deviation : $\sigma = 13.70$

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 5 degrees

$z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76$

For x= 10 degrees

$z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40$

The P-value : $P(-0.76$

$=0.3445783-0.2236273=0.120951\approx0.1210$

In percent , $0.1210\times100=12.10\%$

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

5 0

2 years ago

What is the following product?

umka21 [38]

Answer:

10 sqrt 3.

Step-by-step explanation:

sqrt30 = sqrt10 * sqrt 3

so sqrt30 * sqrt10

= sqrt10 * sqrt10 * sqrt3

= 10 sqrt 3.

5 0

2 years ago

Estimate the limit.

Bezzdna [24]

Answer:

a. 0.25

Step-by-step explanation:

The given expression is

$\lim_{x \to 2} \frac{x-2}{x^2-4}$

Factor the denominator using difference of two squares.

$\lim_{x \to 2} \frac{x-2}{(x-2)(x+2)}$

Cancel out common factors;

$\lim_{x \to 2} \frac{1}{(x+2)}$

We plug in 2 to get

$\lim_{x \to 2} \frac{1}{(x+2)}=\frac{1}{2+2}$

$\lim_{x \to 2} \frac{1}{(x+2)}=\frac{1}{4}=0.25$

3 0

3 years ago

Read 2 more answers