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3 years ago

15

a medical laboratory technician is staining blood film slides with wrights stain . the Wright strain bottle contains 90 milliter

s . She uses 6 male of the stain for each slide . how many slides can stain?

1 answer:

Jet001 [13]3 years ago

4 0

I don't know if I understand the question but I think it is 15

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.

umka21 [38]

Answer:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

$\bar X_1 = 28900$ the sample mean for the salaries of teachers in Indiana

$s_1 = 2300$ the sample deviation for the salary of teachers in Indiana

$n_1 =10$ the sample size from Indiana

$\bar X_2 = 30300$ the sample mean for the salaries of teachers in Michigan

$s_2 = 2100$ the sample deviation for the salary of teachers in Michigan

$n_2 =14$ the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom for this case are:

$df = n_1 +n_2 -2= 10+14-2 =22$

The confidence is 95%so then the significance is $\alpha=0.05$ and the $\alpha/2 =0.025$ , we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2}= 2.07$

And now replacing in the formula for the confidence interval we got:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

6 0

2 years ago

Write the equation of a line that passes through (3.5,0) and is perpendicular to 8y+4x=64.

pantera1 [17]

Answer:

y = 2x + 3.5

Step-by-step explanation:

Step 1: find the slope

8y + 4x = 64

8y = 64 - 4x

Make y the subject of the formula

y = (64 - 4x)/8

y = ( -4x + 64)/8

Separate to get slope

y = -4x/8 + 64/8

y = -x/2 + 8

Slope is the coefficient of x

m = -1/2

Note: if two lines are perpendicular to the other , it is negative reciprocal to each other

m = 2

Using the point slope form equation

y - y1 = m(x - x1)

y - y1 = 2(x - x1)

Substitute the point

( 3.5 , 0)

x1 = 3.5

y1 = 0

y - 3.5 = 2( x - 0)

open the bracket

y - 3.5 = 2x - 0

y = 2x - 0 + 3.5

y = 2x + 3.5

The equation of the line is

y = 2x + 3.5

7 0

2 years ago

−4t + 9 ≥ −7 PLEASE HELPPP

stellarik [79]

Answer:

t is less than or equal to 4

Step-by-step explanation:

7 0

2 years ago

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

KengaRu [80]

Answer:

The 95% confidence interval for the difference of means is (7.67, 16.33).

The estimate is Md = 12.

The standard error is sM_d = 2.176.

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1 (this year's sales), of size n1=36 has a mean of 53 and a standard deviation of 12.

The sample 2 (last year's sales), of size n2=49 has a mean of 41 and a standard deviation of 6.

The difference between sample means is Md=12.

$M_d=M_1-M_2=53-41=12$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12^2}{36}+\dfrac{6^2}{49}}\\\\\\s_{M_d}=\sqrt{4+0.735}=\sqrt{4.735}=2.176$

The degrees of freedom are:

$df=n_1+n_2-1=36+49-2=83$

The critical t-value for a 95% confidence interval and 83 degrees of fredom is t=1.989.

The margin of error (MOE) can be calculated as:

$MOE=t \cdot s_{M_d}=1.989 \cdot 2.176=4.328$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 12-4.328=7.67\\\\UL=M_d+t \cdot s_{M_d} = 12+4.328=16.33$

The 95% confidence interval for the difference of means is (7.67, 16.33).

7 0

2 years ago

The phone company A Fee and Fee has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount o

IgorLugansk [536]

Answer:

y=0.2x+29

Step-by-step explanation:

Given that:

y is the total monthly of the A Fee and Fee plan.

x is the number of monthly minutes used.

If a customer uses 290 minutes, the monthly cost will be $87, we have the pair (290, 87)

If the customer uses 980 minutes, the monthly cost will be $225, this is the coordinate pair (980, 225).

We want to obtain an equation in the form: y=mx+b

First, let us determine the slope, m

Given points (290, 87) and (980, 225):

<u>Slope</u>

$m=\dfrac{225-87}{980-290} =\dfrac{138}{690}=0.2$

Next, we determine the y-intercept, b.

Substituting the pair (290, 87) and m=0.2 in y=mx+b, we obtain

87=0.2(290)+b

b=87-0.2(290)=29

Therefore, our equation in the form y=mx+b is:

y=0.2x+29

8 0

2 years ago