Find the value of x in the triangle shown below.

Solve for x. −23(3x−4)+3x=56

Neporo4naja [7]

Open bracket
-69x+92+3x=56
Collect like term
-66x=-36
Divide both side by -66
x=6\11

7 0

4 years ago

Read 2 more answers

Angie bought 500 identical dresses for $12 500.

tatyana61 [14]

Answer:

10.5%

Step-by-step explanation:

The fraction Angie sold at 130% of the cost price was 350/500 = 0.7. The remaining 0.3 of the dresses were sold at 130%/2 = 65% of the cost price.

Angie's net on the deal was ...

(0.7)(130%) +(0.3)(65%) = 91% +19.5% = 110.5%

where 100% is Angie's cost price. Her profit is this percentage less 100%.

Angie's profit as a percentage of her cost price was 10.5%.

_____

Alternate solution

Obviously, you can work out the cost price of each dress ($25) and the marked up price ($32.50). Then you can work out the revenue for 350 dresses ($11,375) and the revenue from the sale of the remaining 150 dresses at half price ($16.25 × 150 = $2,437.50). Angie's total revenue ($13,812.50) less the cost price is her profit ($1,312.50). Divided by her original cost, this will be her profit: $1,312.50/$12,500 × 100% = 10.5%.

6 0

4 years ago

A=8, b=?, c=16 find the missing side

Ksivusya [100]

Answer:

13.8 I think

Step-by-step explanation:

using the pyrhagorem theorem use the formula c=a2+b2

4 0

3 years ago

Sheila eats 3/4 of a bag of carrots each week, how many bags of baby carrots does she eat in 6 weeks? Write in simplest form

Alla [95]

3/4 x 6 = 18/4 = 4 1/2

she eats 4 1/2 bags of carrots in 6 weeks

8 0

3 years ago

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A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selecti

Andrews [41]

Answer:

(a) Null Hypothesis, $H_0$ : p = 0.50

Alternate Hypothesis, $H_A$ : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

(e) The value of z test statistics is 0.96.

(f) The P-value is 0.1685.

(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) We conclude that the proportion of baby girls is equal to 0.50.

Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 births.

Let p = population proportion of baby girls

(a) So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of baby girls = $\frac{78}{144}$ = 0.54

n = sample of births = 144

So, the test statistics = $\frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }$

= 0.96

The value of z test statistics is 0.96.

(f) The P-value of the test statistics is given by;

P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

= 1 - 0.8315 = 0.1685

(g) Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which we fail to reject our null hypothesis.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

7 0

3 years ago