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murzikaleks [220]
3 years ago
7

In each bouquet of flowers, there are 5 roses and 8 white carnations. Complete the table to find how many roses and carnations t

here are in 3 bouquets of flowers.
Roses: 5,10,15,20,25
Carnations:8,(_),(_),(_),(_)
There are__ roses and __ carnations in the 3 bouquets of flower.
Mathematics
1 answer:
Verizon [17]3 years ago
5 0
40 70 is the right answer ❤️❤️❤️❤️❤️❤️❤️
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I need help now its due right now can anyone help me
cestrela7 [59]

Answer:

(2,-3)

Step-by-step explanation:

I thought you'd actually have to calculate it but I guess they just wanted to give you guys a graph lol

7 0
2 years ago
Answer the question about the image.
Anton [14]

Answer:

4

Step-by-step explanation:

hope this helps and have a great day

3 0
2 years ago
The weights of the fish in a certain lake are normally distributed with a mean of 12 lb and a standard deviation of 12. If 16 fi
erastova [34]

Answer:

If 4 fish are randomly selected, what is the probability that the mean weight will be between 9.6 and 15.6 lb? Round your answer to four decimal places.

6 0
3 years ago
Look at the graph shown below:
ElenaW [278]
So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
%   (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}

\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}

once you get the slope and solve for "y", that'd be the equation of the line.
7 0
3 years ago
The distance between the points (-3,-1) and (-4,0) is the square root of ___?
Galina-37 [17]
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -3}}\quad ,&{{ -1}})\quad 
%  (c,d)
&({{ -4}}\quad ,&{{0}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
6 0
3 years ago
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