All categories

1 year ago

Elliott has some yarn that she wants to use to make hats and scarves. Each hat uses 0.2 kilograms of yarn and

Mathematics

2 answers:

erastova [34]1 year ago

8 0

Answer:

The equation are:

1) h + s = 20

2) 4h = 1s

Step-by-step explanation:

Number of hats Elliott made = h

Number of scarfs Elliott made = s

total number of items Elliott made = 20

h + s = 20 ..[1]

1 hat uses 0.2 kilograms of yarn

1 hat = 0.2 kg of yarn

Then h hats will use = 0.2h kg of yarn

1 scarf uses 0.1 kilograms of yarn

1 scarf= 0.1 kg of yarn

Then s scarfs will use = 0.1s kg of yarn

She wants to twice as much as yarn for scarves for hats:

2 × (0.2h kg of yarn) = 0.1s kg of yarn

0.4h = 0.1s

4h = 1s...[2]

The equation are:

1) h + s = 20

2) 4h = 1s

RUDIKE [14]1 year ago

4 0

Answer:

$\left\{\begin{array}{l}h+s=20\\ \\0.1s=2\cdot 0.2h\end{array}\right.$

Step-by-step explanation:

Let h be the number of hats Elliott makes and s be the number of scarves she makes. She wants to make a total of 20 items, so

$h+s=20$

Each hat uses 0.2 kilograms of yarn, then h hats use 0.2h kg of yarn. Each scarf uses 0.1 kilograms of yarn, then s scarves use 0.1s kg of yarn. Elliott wants to use twice as much yarn for scarves as for hats, so

$0.1s=2\cdot 0.2h$

You get the system of two equations:

$\left\{\begin{array}{l}h+s=20\\ \\0.1s=2\cdot 0.2h\end{array}\right.$

You might be interested in

You know that the current exchange rate for US dollars (USD) to Indian rupees (INR) is 1:45.85, that the exchange rate for India

dem82 [27]

Answer:

Step-by-step explanation:

the answer on edgeunity

3 0

2 years ago

Read 2 more answers

Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

-------

Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$

5 0

2 years ago

Uma pessoa realizou uma compra que foi financiada em três parcelas mensais e iguais a R$ 350,00, o financiamento foi realizado s

lisabon 2012 [21]

Answer: Amount of purchase would be $393.70.

Step-by-step explanation:

One person made a purchase that was financed in three monthly installments and equal to $ 350.00, the financing was made at the compound interest rate of 48% a.a. Determine the amount of the purchase.

Since we have given that

Principal = $350.00

Rate of compounded monthly = 48%

Number of months = 3

So, the amount of purchase would be

$Amount=350(1+\dfrac{0.48}{12})^3\\\\Amount=350(1+0.04)^3\\\\Amount=350(1.04)^3\\\\Amount=\$393.70$

Hence, Amount of purchase would be $393.70.

5 0

1 year ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago

At a pond there were 24 ducks swimming. The ratio of ducklings to adult ducks is 5:1. How many ducklings were swimming at the po

grigory [225]

Twenty ducklings were swimming at the pond

7 0

1 year ago

Read 2 more answers