What is the value of x in the equation −6 + x = −2? (5 points) a 8 b 4 c −4 d −8

logan trades 50 pennies for nickels he gets a nickel for each group of 5 pennies what equation can be used to find how many nick

V125BC [204]

The equation that can be used to find how many nickels logan uses is: 50÷5.

<h3>Equation to find the number of nickels</h3>

Given:

Total pennies=50

Logan got a nickel of 5 pennies=1

Hence:

Number of nickels=1÷5×50

Number of nickels=50÷5

Number of nickels=10 nickels

Therefore the equation that can be used to find how many nickels logan uses is: 50÷5.

Learn more about Equation to find the number of nickels here:brainly.com/question/10704332

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4 0

1 year ago

Kyle received the following course grades for 10 (three-credit) courses during his freshman year:

timofeeve [1]

Answer:

(a) The average grade point is 2.5.

(b) The relative frequency table is show below.

(c) The mean of the relative frequency distribution is 0.3333.

Step-by-step explanation:

The given data set is

4, 4, 4, 3, 3, 3, 1, 1, 1, 1

(a)

The average grade point is

$Average=\frac{\sum x}{n}$

$Average=\frac{4+4+4+3+3+3+1+1+1+1}{10}$

$Average=2.5$

Therefore the average grade point is 2.5.

(b)

$\text{Relative frequency}=\frac{\text{Frequency of number}}{\text{Total frequency}}$

The relative frequency table is show below:

x f Relative frequency

4 3 $\frac{3}{10}=0.3$

3 3 $\frac{3}{10}=0.3$

1 4 $\frac{4}{10}=0.4$

(c)

Mean of the relative frequency distribution is

$Average=\frac{0.3+0.3+0.4}{3}$

$Average=0.3333$

Therefore the mean of the relative frequency distribution is 0.3333.

4 0

2 years ago

Need help anyone can help giving brain liest

trapecia [35]

Answer:

14 gel pens/minute

Step-by-step explanation:

divide the amount by time.

210/15 = 14

6 0

2 years ago

Read 2 more answers

Given the explicit rule, f(n) = 2n - 3, generate the first four terms in the sequence.

Vikki [24]

Answer: -1, 1, 3, 5

Step-by-step explanation:

F(n)= 2n-3

F(1)=2(1)-3= -1

F(2)=2(2)-3= 1

F(3)=2(3)-3= 3

F(4)=2(4)-3=5

5 0

2 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago