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2 years ago

Which group of ordered pairs are on the line given by the equation 5x – 2y = 6?

1 answer:

3 0

<h2>There are infinite ordered pairs are on the line because it has infinite solutions. I think but i'm not very sure.</h2>

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Solving Exponential Equations (lacking a common base)<br> 4^x-5=8

avanturin [10]

Answer:

x= 6 1/2 or 6.5

Step-by-step explanation:

4^x-5 =8

2^2(x-5) =2^3

since the base is equal, the powers are equal

2(x-5) = 3

2x - 10 = 3

2x = 3 + 10

2x = 13

2x/2 = 13/2

x = 6 1/2 or 6.5

8 0

2 years ago

Read 2 more answers

HELP ME PLEASEEEEEEEEE

natima [27]

I think its the last one, Aidan can expect to save about $27.52 each week. Hope I was correct

6 0

2 years ago

A circle graph has five

Norma-Jean [14]

Answer:

the missing section is 49%

Step-by-step explanation:

11 + 18 + 15 + 7= 51

51 + 49 = 100

add the percentages together and you get 100%

4 0

1 year ago

There was 7L of punch for the birthday party. How many mL of punch was<br> there?

Mnenie [13.5K]

Answer:

7000 ml

Step-by-step explanation:

hope this helps!!!!!!!!!

3 0

2 years ago

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

7 0

2 years ago