Please I really need help

valina [46]

X=8 i did mental math but 56x-16-x=424
                                            +16      +16
55x=440
   /55 /55
                                                x=8

3 0

3 years ago

Alonzo wrote the equation x=2+4y-5 solve for y

kati45 [8]

hi

if x= 2+4y-5 then x-2 +5 = 4y so y = (x-2+5) /4 = ( x+3) /4

conclusion y = (x+3)/4

3 0

3 years ago

A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

Natali [406]

Answer:

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.17$ represent the mean for the sample 1 (25 mil film)

$\bar X_{2}=1.04$ represent the mean for the sample 2 (20 mil film)

$s_{1}=0.11$ represent the sample standard deviation for the sample 1

$s_{2}=0.09$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size selected for 1

$n_{2}=8$ sample size selected for 2

$\alpha=0.05$ 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 reducing the film thickness increases the mean speed of the film, the system of hypothesis would be:

Null hypothesis: $\mu_{1} \leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{1.17-1.04}{\sqrt{\frac{0.11^2}{8}+\frac{0.09^2}{8}}}}=2.587$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a one sided test the p value would be:

$p_v =P(t_{(14)}>2.587)=0.0108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we have enough evidence to reject the null hypothesis on this case and the 25 mil film have a mean greater than the 20 mil film so then the claim is not appropiate

4 0

3 years ago

HELP ME HELP ME HELP Help help

alina1380 [7]

Answer:

<

Step-by-step explanation:

-3 4/5 is a small number than -3 1/5 because it is a greater negative number, which means it is farther away from 0 on a number line Ex:

5 4 3 2 1 0 -1 -2 -3 -4 -5

6 0

3 years ago

Read 2 more answers

Mrs. Rifkin had 84 oranges. 24 of her oranges were rotten. The rest were either ripe or unripe. There were 18 more ripe oranges

Natalka [10]

Answer:

13:8:7

Step-by-step explanation:

84 - 24 = 60

60 oranges were either ripe or unripe.

18 more ripe oranges than unripe oranges.

x + 18 = ripe oranges

x = unripe oranges

x + 18 + x = 60

2x + 18 = 60

2x = 42

x = 21

21 are unripe oranges

21 + 18 = 39

39 oranges are ripe.

Ratio of the number of ripe oranges to the number of rotten oranges to the number of unripe oranges.

39:24:21

13:8:7

4 0

3 years ago