PLZ HELP IM IN A HURRY!

Can someone give me the answers plz:)

Alborosie

Remark

There is no short way to do this problem and no obvious way to get the answer other that to solve each part.

Solve

A

$\dfrac{x + 1.6}{2} =\text{x + 0.1}$ Multiply by 2

x + 1.6 = 2(x + 0.1) Remove the brackets

x + 1.6 = 2x + 0.1*2

x + 1.6 = 2x + 0.2 Subtract x from both sides

1.6 = x + 0.2 Subtract 0.2 from both sides

1.6 - 0.2 = x

1.4 = x

Circle A

B

Subtract 2x from both sides.

3x - 2x = 1.4

Circle B

C

Remove the brackets.

4x + 6 = 2x - 6 Add 6 to both sides

4x + 12 = 2x Subtract 4x from both sides.

12 = -2x Divide by - 2

12/-2 = x

x = - 6 Don't circle C

D

I'm going to be very scant in my solution of this. You can fill in the steps.

3x = 4.2

x = 4.2/3

x = 1.4

Circle D

5 0

4 years ago

Read 2 more answers

Please Help. ASAP!!

Ratling [72]

Answer:

Equation: x + (x+ 1) = 67

Step-by-step explanation:

We want to find two consecutive integers.

Let them be labeled as x and x + 1

We want x + (x + 1) = 67

x is the smaller number

x + 1 is the largest

We can solve for x.

x + (x + 1) = 2x + 1 = 67

2x = 66

x = 33

The largest integer is 34

3 0

3 years ago

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago

on a map, the length of a nature center Trail is 6.5 cm. If the scale is 2 cm to 10 km oh, what is the actual Trail? BRAINLY

mash [69]

Answer:

32.5 km

Step-by-step explanation:

6.5 / 2 = 3.25

3.25 x 10 = 32.5 km

Hope that helps!

6 0

3 years ago

Read 2 more answers

In a recent year, grade 8 Washington State public school students taking a mathematics assessment test had a mean score of 281 w

sweet-ann [11.9K]

Answer:

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

Step-by-step explanation:

Step(i):-

Mean of the Population (μ) = 281

Standard deviation of the Population = 34.4

Let 'X' be a random variable in Normal distribution

Given X = 290

$Z = \frac{x -mean}{\frac{S.D}{\sqrt{n} } } = \frac{290-281}{4.44} = 2.027$

Step(ii):-

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = P( Z > 2.027)

= 0.5 - A ( 2.027)

= 0.5 - 0.4783

= 0.0217

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

6 0

3 years ago