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3 years ago

22x+25-19=180 what does x equal

Mathematics

2 answers:

siniylev [52]3 years ago

8 0

Answer:

x = 87/11 = 7.909

Step-by-step explanation:

Step 1 :

Pulling out like terms :

1.1 Pull out like factors :

22x - 174 = 2 • (11x - 87)

Equation at the end of step 1 :

Step 2 :

Equations which are never true :

2.1 Solve : 2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2 Solve : 11x-87 = 0

Add 87 to both sides of the equation :

11x = 87

Divide both sides of the equation by 11:

x = 87/11 = 7.909

One solution was found :

x = 87/11 = 7.909

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Softa [21]3 years ago

6 0

Answer:7.9 approximately 8

Step-by-step explanation:

22x + 25 -19 =180

22x =180 -25 +19

22x = 180 - 6

22x = 174

x =174/22

x = 7.9 or approximately 8

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Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

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

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

= [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

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

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50 {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

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

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

n = sample of babies = 400

So, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

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Trish and Aria are walking to raise money for the school library. Trish has walked 38 miles and walks 10 miles each week. Aria has walked 30 miles and walks 12 miles each week

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Step-by-step explanation:

From the information given:

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Suppose X be the number of accidents involved by a single-vehicle.

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$X \sim Binom (15,0.7)$

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