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Ivahew [28]
2 years ago
7

This is confusing me! Please help​

Mathematics
1 answer:
podryga [215]2 years ago
8 0

Answer:

£2.85

Step-by-step explanation:

Lucy paid this amount she used six coin remember you have to calculate them to get the 85 and u can't use £1

three different ways she could have paid

50+20+10+5=85(£2.85)

50+10+10+10+5=85(£2.85)

20+20+20+10+10+5=(£2.85)

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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

4 0
3 years ago
In a randomized controlled trial in Kenya, insecticide treated bednets were tested as a way to reduce malaria. Among 343 infants
Dovator [93]

Answer:

Step-by-step explanation:

Given that in a randomized controlled trial in Kenya, insecticide treated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria.

H0: p1=p2

H1: p1 <p2

(one tailed test)

p1 = 15/343:p2 =27/294:

Difference 4.813 %

95% CI 0.9160% to 9.0217%

Chi-squared 5.947

DF  1

Significance level P = 0.0073

Since p <0.01 we reject null hypothesis.

4 0
3 years ago
At the football game, Joe carried the ball four different times. He had a loss of 15, a gain of 2, a loss of 5, and a gain of 7.
juin [17]
15+2=17-5=12+7=20 B: -20 yards
4 0
2 years ago
What does it mean when a line changes and you translated
dsp73
Translation means you are shifting the line either up or down.
3 0
3 years ago
18 minutes left hwhsusbe
Pepsi [2]

Answer:

so its D

Step-by-step explanation:

a is wrong becuase it does have 6

b is wrong becuase you make into 6 triangles

c is wrong becuase it does add up to 180 degrees

6 0
3 years ago
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