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Ted has a fixed amount of money in his budget for holiday gifts. He decides that his wife and his mother will receive gifts of e

vredina [299]

Answer would be A- 6x+20 his siters gift is x, (x)6=6 x
moms gift and wifes gift are each a (2x) than his two (2x) and 2(x's) = 6x.
2x+2x+x+x=6x, so now 6x+20 equals total ammount spent (total budget)

4 0

3 years ago

What is the value of x for the givin equation 4-2(x+7)=3(x+5)

Elenna [48]

Answer:

-5

Step-by-step explanation:

$4-2x-14=3x+15\\-2x-10=3x+15\\-5x=25\\x=-5$

5 0

3 years ago

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Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for stro

tigry1 [53]

Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation

n=2700 represent the random sample taken

X represent the people in England who are deficient in Vitamin D

$\hat p=0.2$ estimated proportion of England people who are deficient in Vitamin D

$\alpha=0.02$ represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And replacing into the confidence interval formula we got:

$0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182$

$0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218$

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

3 0

3 years ago

Brian has reduced his cholesterol level by 18% after his last check up. If his original level was 220, what is his approximate

Drupady [299]

Answer:

B

Step-by-step explanation:

multiply 18% and 220 which is 39.6

subtract 39.6 from 220 which is 180.4

simplify

3 0

3 years ago

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Triangle DEF (not shown) is similar to ABC shown, with angle B congruent to angle E and angle C congruent to angle F. The length

Bezzdna [24]

Answer:

Area of ΔDEF is $45\ cm^2$ .

Step-by-step explanation:

Given;

ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Length of AB = $2\ cm$ and

Length of DE = $6\ cm$

Area of ΔABC = $5\ cm^2$

Solution,

Since, ΔABC and ΔDEF is similar and ∠B ≅ ∠E.

Therefore,

$\frac{Area\ of\ triangle\ 1}{Area\ of\ triangle\ 2} =\frac{AB^2}{DE^2}$

Where triangle 1 and triangle 2 is ΔABC and ΔDEF respectively.

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

$\frac{5}{Area\ of\ triangle\ 2} =\frac{2^2}{6^2}\\ \frac{5}{Area\ of\ triangle\ 2}=\frac{4}{36}\\ Area\ of\ triangle\ 2=\frac{5\times36}{4} =5\times9=45\ cm^2$

Thus the area of ΔDEF is $45\ cm^2$ .

4 0

3 years ago

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