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3 years ago

I WILL GIVE BRAINLY

Mathematics

1 answer:

Alenkinab [10]3 years ago

6 0

Answer:

5(6)+0.65y

Step-by-step explanation:

$5(6hrs)+65÷100=0.65

=5(6)+0.65y

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Find the missing angle measures of the following figure

Lisa [10]

Answer:

1 = 68

2 = 68

3 = 125

Step-by-step explanation:

Hey there!

To find angle 1 all we have to do is subtract the given angles from 180 because remember all of the angles in a triangle add up to 180

180-73-39=68

1=68

angle 2 is vertical to angle 1 therefore they are congruent angles so

2 also equals 68

So we know that the interior angles in a 4 sided shape add up to equal 360

so to find 3 we subtract the given angles from 360

360-90-77-68=125

so 3 = 125

6 0

3 years ago

Can you help me calculate the angles from this kite in the attachment?<br><br><br>AY =, XZ =,

Setler [38]

Answer:

AY=8, XZ=20

Step-by-step explanation:

WY=WA+AY, plug in values: 14=6+AY, subtract 6 from both sides: 8=AY, AY=8cm

ZA=10cm and WY bisects XZ at A, so XA=ZA and XZ = XA+ZA = 2ZA =2*10 = 20 cm

8 0

3 years ago

Desire <br> Help me with this quistion

aleksandrvk [35]

Answer:

The value of n is 11 feet.

Step-by-step explanation:

Given that the area of rectangle from is A = length×width so we have to substitute the variables into the formula in order to find n :

$area = length \times width$

Let area = 33,

Let length = n,

Let width = 3,

$33 = n \times 3$

$3n = 33$

$n = 33 \div 3$

$n = 11 \: feet$

8 0

3 years ago

a suit cost $95 Barry has a discount coupon for 10% off and the city has a 4.5 percent sales tax what is the final cost of the s

SIZIF [17.4K]

SO the suit cost $95 and it has a discount of 10%
so 95*10/100=9.5
so 10% of 95 is 9.5
subtract 9.5 from 95 and that equals 85.5
then you have to add 4.5 percent sales tax
so 85.5+ 4.5%= 89.347
So the final cost would be $89.34

3 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago