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3 years ago

-4b-5+2b=10 Help me solve this. I'm thinking way to hard.

1 answer:

8 0

Isolate the variables. You want to get the five on the same side as the ten, so, you add five to both sides.

-4b+2b=15

-2b=15
b= -7.5

I believe. You may want to double check my work though.

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Ms. Barclay orders birthday cupcakes for the month of June from an online vendor. Each cupcake costs $1.25 and there is a one-ti

artcher [175]

Answer:

Ms. Barclay ordered 9 cupcakes.

Step-by-step explanation:

$1.25x9=11.25

11.25+3.25=$14.50

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3 years ago

K+2 over 3= -7 <br> How do u solve this problem

harkovskaia [24]

K+2/3=-7
K+2=-7x3
K+2=-21
K=-19

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3 years ago

Is 63.90 greater than 63.990

djverab [1.8K]

Answer:

63.90 is greater than63.990

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2 years ago

Read 2 more answers

Maelin raised $50 plus $3 for each mile, m, that she walked in a fund-raiser walk. Which expression represents the situation? Ch

zaharov [31]

Answer:

50+3m=x

Step-by-step explanation:

Maelin gave $50

$3 per mile

M= miles

50+3m=x

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2 years ago

Read 2 more answers

An insurance company selected a random sample of 500 clients under 18 years of age and found that 180 of them had had an acciden

Butoxors [25]

Answer:

a) The pooled proportion is p=0.3.

b) P-value = 0.000078

c) Lower bound = 0.0556

d) Upper bound = 0.1644

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the accident proportions differ between the two age groups .

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=500 has a proportion of p1=0.36.

$p_1=X_1/n_1=180/500=0.36$

The sample 2, of size n2=600 has a proportion of p2=0.25.

$p_1=X_1/n_1=150/600=0.25$ .

The difference between proportions is (p1-p2)=0.11.

$p_d=p_1-p_2=0.36-0.25=0.11$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{180+150}{500+600}=\dfrac{330}{1100}=0.3$

The standard error for the difference between proportions can now be calculated as:

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.3*0.7}{500}+\dfrac{0.3*0.7}{600}}\\\\\\s_{p1-p2}=\sqrt{0.00042+0.00035}=\sqrt{0.00077}=0.0277$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.11-0}{0.0277}=\dfrac{0.11}{0.0277}=3.964$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t>3.964)=0.000078$

As the P-value (0.000078) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the accident proportions differ between the two age groups.

If we want to calculate the bounds of a 95% confidence interval, we start by calculating the margin of error.

For a 95% CI, the critical value for z is z=1.96.

Then, the margin of error is:

$MOE=z \cdot s_{p1-p2}=1.96\cdot 0.0277=0.0544$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.11-0.0544=0.05561\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.11+0.0544=0.16439$

The 95% confidence interval for the population mean is (0.0556, 0.1644).

5 0

3 years ago