4. Carole worked her 40 regular hours last week plus 5 overtime hours at the time-and-a-half rate.

Someone please help me I’m really struggling I will give brainliest

viktelen [127]

Answer:

SA ≈ 1134 cm²

General Formulas and Concepts:

Math

Pi π = 3.14...

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Radius: r = d/2
Surface Area of a Sphere: SA = 4πr²

Step-by-step explanation:

Step 1: Define

d = 19 cm

Step 2: Find SA

Substitute [R]: r = 19 cm/2
Divide: r = 9.5 cm
Substitute [SAS]: SA = 4π(9.5 cm)²
Exponents: SA = 4π(90.25 cm²)
Multiply: SA = 361π cm²
Multiply: SA = 1134.11 cm²
Round: SA ≈ 1134 cm²

8 0

3 years ago

While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who en

AVprozaik [17]

Answer:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

$p_v =P(Z>0.0057)=0.4977$

The p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

Step-by-step explanation:

1) Data given and notation

$X_{M}=23$ represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

$X_{W}=8$ represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

$n_{M}=66$ sample of male selected

$n_{W}=23$ sample of demale selected

$p_{M}=\frac{23}{66}=0.34848$ represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

$p_{W}=\frac{8}{23}=0.34782$ represent the proportion of women with red/green color blindness

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:

Null hypothesis: $p_{M} \leq p_{W}$

Alternative hypothesis: $p_{M} > p_{W}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}$ (1)

Where $\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057$

4) Statistical decision

Using the significance level provided $\alpha=0.05$ , the next step would be calculate the p value for this test.

Since is a one side right tail test the p value would be:

$p_v =P(Z>0.0057)=0.4977$

So the p value is a very high value and using the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.

4 0

3 years ago

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago

Lexi created 828 bracelets this last school year. If she made an equal amount for each of the 9 months, how many bracelets did L

shusha [124]

Answer:

92 bracelets a month

Step-by-step explanation:

Duvide 828 by 9

5 0

3 years ago

In his boat, Sheldon can travel 45 mi downstream in the same time that it takes to travel 9 mi upstream. If the rate of the curr

Nataliya [291]

Answer:

9 mph

Step-by-step explanation:

-let x be the speed of current and t be time. The speed equation for both directions can then be represented as:

$Speed=\frac{Distance}{Time}\\\\\#Upstream\\(u-6)t=9\\\\\#Donwstream\\\\(u+6)t=45$

#Since t is equal in both, we can do away with t.

#We the divide the downstream equation by the upstream equation as:

$\frac{u+6}{u-6}=\frac{45}{9}\\\\(u+6)=5(u-6)\\\\u+6=5u-30\\\\4u=36\\\\u=9$

Hence, the boat's speed in still water is 9 mph

7 0

3 years ago