PLEASE HELP QUESTION IN PHOTO WILL GIVE BRAINLIST!

Equation A: a=5b + 1

lukranit [14]

ANSWER:

a = 11

b = 2

Explaination:

We have two equations:

a = 5b + 1 .....................(1)

a = 3b + 5.....................(2)

Here we use substitution method in order to solve the set of equations.

put a=3b+1 to equation (1)

3b + 5 = 5b + 1

Solve for b

5 = 5b -3b + 1

5-1 =2b

4 = 2b

b = 2

Now put the value of b = 2 into the equation (1) or (2)

a = 5b + 1

a = 5(2) + 1

a = 10 + 1

a = 11

5 0

4 years ago

A graph that uses bars of various heights to represent the frequencies.

Sunny_sXe [5.5K]

A graph that uses bars of various heights to represent the frequencies is a <u>Histogram</u>

A histogram is an approximate representation of the distribution of numerical data. The term was first introduced by Karl Pearson. To construct a histogram, the first step is to "bin" (or "bucket") the range of values—that is, divide the entire range of values into a series of intervals—and then count how many values fall into each interval.

Therefore, a graph that uses bars of various heights to represent the frequencies is a <u>Histogram</u>

5 0

2 years ago

Read 2 more answers

If a car is moving at a speed of 30 miles per hour, how many miles

Ne4ueva [31]

Answer:

90 miles

Step-by-step explanation:

There's more than one way in which to do this. I have chosen to convert '180 minutes' to '3 hours.'

Since distance = rate times time, we have here:

distance traveled in 180 minutes = (30 mph)(3 hr) = 90 miles

4 0

3 years ago

An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining.

Tanya [424]

Answer:

98% Confidencce Interval is ( 3030.6, 7467.4 )

Step-by-step explanation:

Given that:

Sample size $n_1 =$ 71

Sample size $n_2 =$ 31

Sample mean $\overline x_1 =$ 41628

Sample mean $x_2 =$ 36,379

Population standard deviation $\sigma_1$ = 4934

Population standard deviation $\sigma_2 =$ 4180

At 98% confidence interval level, the level of significcance = 1 - 0.98 = 0.02

Critical value at $z_{0.02/2} = 2.33$

The Margin of Error = $z \times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2} }$

= $2.33 \times \sqrt{\dfrac{4934^2}{71}+\dfrac{4180^2}{31} }$

= $2.33 \times \sqrt{\dfrac{24344356}{71}+\dfrac{17472400}{31} }$

= $2.33 \times \sqrt{906504.06 }$

= 2218.40

The Lower limit = $( \overline x_1 - \overline x_2) - (Margin \ of \ error)$

= ( 41628 - 36379 ) - ( 2218.40)

= 5249 - 2218.40

= 3030.6

The upper limit = $( \overline x_1 - \overline x_2) + (Margin \ of \ error)$

= ( 41628 - 36379 ) + ( 2218.40)

= 5249 + 2218.40

= 7467.4

∴ 98% Confidencce Interval is ( 3030.6, 7467.4 )

5 0

4 years ago

Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra

Kobotan [32]

<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = sample mean

n= sample size

s= sample standard deviation

$t_{\alpha/2}$ = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , $\overline{x}=22.4$

$\alpha=1-0.99=0.01$

A) Critical t value for $\alpha/2=0.005$ and degree of 18 freedom

$t_{\alpha/2}$ = 2.8784

B) Required confidence interval:

$22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)$

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0

4 years ago