The topping are nuts hot fudge caramel and sprinkles

Which one is it ???????

hram777 [196]

15 is the correct answer

4 0

2 years ago

How to add 3/5 and 1/10 in simplest form

dolphi86 [110]

Find a common denominator and then add.
3/5=6/10
6/10+1/10=7/10

5 0

2 years ago

Read 2 more answers

What is the solution to the equation below? √x - 4 = -2 O A. X=& O B. X=4 X = 4 C. x = 6 D. X = 2

saul85 [17]

Answer:

x = 4

Explanation:

Given the expression;

$\sqrt[]{x}-4\text{ = -2}$

Add 4 to both sides

$\begin{gathered} \sqrt[]{x}-4+4\text{ = -2+4} \\ \sqrt[]{x}=\text{ 2} \end{gathered}$

Square both sides

$\begin{gathered} (\sqrt[]{x})^2=2^2 \\ x\text{ = 4} \end{gathered}$

Hence the value of x is 4

$undefined$

5 0

11 months ago

Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case

olga nikolaevna [1]

Answer:

(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> = Mike will show up.

Given:

$P(D^{c})=0.55\\P(M^{c})=0.45$

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

$P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25$

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

$=1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75$

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

$P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25$

Thus, the probability that neither Dave nor Mike will show up is 0.25.

6 0

3 years ago

Read 2 more answers

Question in attachment

kiruha [24]

Answer:

1. a= 7, A = 49 degrees

Side A = 7 , Side B = 24, Side C = 25

A= 49 degrees, C = 90 degrees , B=41

2. A= 4, B = 6.9, C=8

A= 22 degrees, B = 68, C= 90

3. A= 7, B = 14.4, C=16

A= 45 , B=45 , C=90

6 0

3 years ago