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3 years ago

7

SOME ONE HELP MEEE ITS TIMED

2 answers:

Alex Ar [27]3 years ago

4 0

Answer:

b

Step-by-step explanation:

Solnce55 [7]3 years ago

3 0

I cant see the other graphs

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Help with this?<br>find the surface area

expeople1 [14]

Answer:

49.

7×7=49.

Step-by-step explanation:

i did this. hope it helps

6 0

3 years ago

PLEASE HELP ASAP!!!!!

Wittaler [7]

Answer:

$sec(x)$

Step-by-step explanation:

$cos(x)+cos(x)tan^2(x)$

Multiply cos(x) by 1:

$cos(x)*1+cos(x)tan^2(x)$

Factor out cos(x):

$cos(x)(tan^2(x)+1)$

Apply the pythagorean identity ( $tan^2(x)+1=sec^2(x)$ ):

$cos(x)sec^2(x)$

Rewrite $sec^2(x)$ :

$cos(x)(\frac{1}{cos(x)})^2 \\cos(x)\frac{1}{cos^2(x)}$

Cancel cos(x):

$\frac{1}{cos(x)}$

Rewrite:

$sec(x)$

5 0

4 years ago

The graph below represents the amount of change Jaxon would receive, y, if he bought a item with a cost of x. Which equation rep

harina [27]

Answer:

1. y = 10 - x

Step-by-step explanation:

Because he starts off with 10 dollars, and x is the amount he spends on an item, y is showing the change that he would get from paying with a 10 dollar bill, so it's what's left over from that 10 dollars.

3 0

3 years ago

Free points everyone<br><br><img src="https://tex.z-dn.net/?f=10%20%2B%2062%20%5Cdiv%202" id="TexFormula1" title="10 + 62 \div 2

Brut [27]

41 is the right answer

8 0

3 years ago

Read 2 more answers

Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They

ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

$P = 2$

$G = 3$

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

This gives:

$P(G_1\ and\ G_2) = P(G_1) * P(G_2)$

$P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}$

$P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}$

$P(G_1\ and\ G_2) = \frac{3}{10}$

$P(P_1\ and\ P_2) = P(P_1) * P(P_2)$

$P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}$

$P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}$

$P(P_1\ and\ P_2) = \frac{1}{10}$

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

$P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)$

$P(Same) = \frac{3}{10} + \frac{1}{10}$

$P(Same) = \frac{3+1}{10}$

$P(Same) = \frac{4}{10}$

$P(Same) = \frac{2}{5}$

8 0

3 years ago