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Thepotemich [5.8K]
3 years ago
5

At a sandwich shop, Diana can select from 2 types of bread and 5 types of meat. if she randomly selects 1 type of bread and 1 ty

pe of meat, how many possible choices does she have?
Mathematics
2 answers:
WITCHER [35]3 years ago
5 0
She has 10 choices :)
Elena L [17]3 years ago
5 0

Answer:

10 Choices

Step-by-step explanation:

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Solve the inequalities by graphing.
dlinn [17]

Answer:

The 7th graph

blue, green, yellow

3 0
2 years ago
Solve the equation. 8x2 – 4 = 28
Flura [38]
Hello,

Your answer would be 2, -2.

8x^2 - 4 = 28

8x^2-4+4=28+4

8x^2 = 32

\frac{8x^2}{8}=  \frac{32}{8}  


x^2 = 4

x = +/4 

x = 2 or x = -2

2 , -2 would be the answer! :)


Mark Brainliest if this helped you :)
4 0
2 years ago
Read 2 more answers
If p and q are both true, then which of the following statements has the same truth-value as ~p → q?
olganol [36]

If p and q are both true, then

\neg p \implies q

is an implication of the form

F \implies T

which is true, because every implication starting with false is true, i.e.

F \implies T = T,\quad F \implies F = T

So, we're looking for an expression evaluating to true. Let's see what we have:

A) is an AND proposition. Logical AND is true only if both parts are true. So, you have

\neg P \land \neg Q = F \land F = F

So it's not the right option.

B) is an OR proposition. Logical OR is true whenever one of the two parts is true. So, you have

\neg P \lor\neg Q = F \lor F = F

So it's not the right option.

C) is again an AND proposition. You have

P \land \neg Q = T \land F = F

So this is not the right option.

D) Finally, the last one is again an implication, and again it starts with false:

\neg Q \implies P = F \implies T = T

So this is true, and thus is the correct option.

8 0
2 years ago
Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to f
BabaBlast [244]

Answer:

153 times

Step-by-step explanation:

We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14

Width = 0.14

ME = \frac{width}{2}

ME = \frac{0.14}{2}

ME = 0.07

ME\geq z \times \sqrt{\frac{\widecap{p}(1-\widecap{p})}{n}}

use p = 0.5

z at 95.8% is 1.727(using calculator)

0.07 \geq 1.727 \times \sqrt{\frac{0.5(1-0.5)}{n}}

\frac{0.07}{1.727}\geq sqrt{\frac{0.5(1-0.5)}{n}}

(\frac{0.07}{1.727})^2 \geq \frac{0.5(1-0.5)}{n}

n \geq \frac{0.5(1-0.5)}{(\frac{0.07}{1.727})^2}

n \geq 152.169

So, Option B is true

Hence  we have to flip 153 times the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head

6 0
2 years ago
I really need help tyy
emmainna [20.7K]
Find the greatest common factor (gcf) in each grade between the girls and boys
in 6th grade the gcf is 8
64/8=8  72/8=9
the largest number of groups that can be made equally would be 8 groups of 8 girls and 8 boys
Hope this helps!!
6 0
3 years ago
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