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3 years ago

A baby gains 11 pounds in its firat year of life. The baby gained 4 pounds during the first four months and

Mathematics

1 answer:

inysia [295]3 years ago

8 0

Answer:

I think it's 3.5

Step-by-step explanation:

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Caleb is making lemonade and cookies for the school bake sale. He has 10 cups of

Goryan [66]

Answer:

He can make only one batch of cookies.

Step-by-step explanation:

Just by mental math you can say he can make maximum one batch of cookies.

Total sugar = 10 cups

sugar used for lemonade = 2 cups

leftover sugar = 8 cups

For 1 batch of cookie sugar needed = 11/2

which means 5 and 1/2 cups

And he's just left with 8 cups of sugar after lemonade

8 - 5 1/2 = 2 1/2 which is not enough for 2nd batch

That means the maximum number of cookies batches is 1.

Another way to solve it by inequality:

2+Il/2 X < 10

11/2 X < 8

11X < 16

X < 1 6/11

So he can make only one batch of cookies.

3 0

3 years ago

Which type of graph would be best for showing the average yearly bonus amounts for employees in different departments of an IT c

nika2105 [10]

I believe it is C. a histogram
(if it’s not then sorry)

5 0

3 years ago

which of the following is the equation of the line that passes though the point (-2,1) and has a slope of 4

Ierofanga [76]

An equation that goes through (-2, 1) and has a slope of 4 is y = 4x + 9.

You can find this by looking for the y-intercept (b) by using the slope (m), the point and slope intercept form. The work is below for you.

y = mx + b

1 = 4(-2) + b

1 = -8 + b

9 = b

Now we can use that and the slope to create the equation y = 4x + 9

4 0

3 years ago

What is h(x) = 7 + 10x + x2 written in vertex form? h(x) = (x – 25)2 – 18 h(x) = (x – 5)2 + 32 h(x) = (x + 5)2 – 18 h(x) = (x +

muminat

Answer:c

Step-by-step explanation:

6 0

2 years ago

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = -1 000.00

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

3 years ago