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1 year ago

It’s not 0.615 , or 0.75 or the one shown in the picture

Mathematics

1 answer:

Dafna1 [17]1 year ago

6 0

$2.40714$

1) Let's evaluate that expression, given that a=4.9, b=-7, and c=-0.5

$\begin{gathered} a^3\div b^2+c^3 \\ (4.9)^3\div(-7)^2+(-0.5)^3= \\ \frac{4.9^3}{\left(-7\right)^2+\left(-0.5\right)^3} \\ \\ \frac{4.9^3}{48.875} \\ \\ \frac{117.649}{48.875} \\ \\ 2.40714 \end{gathered}$

Note that we have rewritten it as a fraction so that we can easily operate. Also, we have applied here the PEMDAS order of operations, prioritizing the exponents.

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Can someone plz help me?

earnstyle [38]

Answer:

14.8

Step-by-step explanation:

4 0

2 years ago

ranslate the sentence into an equation using n as the unknown number. Then solve the equation for n. Two times a number increase

joja [24]

Hi there!

Let's break this down:
Two times a number = 2n
Increased by 5 = + 5
is 15 - = 15

Total equation:
2n + 5 = 15

Solving:
2n + 5 = 15
2n = 10
n = 5

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

4 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

3 years ago

Please help! Consider the given circle with the shaded sector xy and central angle 225. The circumference is 30 units

OleMash [197]

$\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=30\pi \end{cases}\implies 30\pi =2\pi r\implies \cfrac{30\pi }{2\pi }=r\implies \boxed{15=r} \\\\[-0.35em] ~\dotfill$

$\textit{arc's length}\\\\ s = \cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=15\\ \theta =225 \end{cases}\implies \widehat{XY}=\cfrac{(225)\pi (15)}{180}\implies \boxed{\widehat{XY}\approx 58.90} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad A=\pi (15)^2\implies A=225\pi \implies \boxed{A\approx 706.86}$

6 0

2 years ago

Emma made a scale model of the Sam Houston's monument. The monument's actual height is 67 ft tall. Emma is using a scale where 1

Kamila [148]

Answer: 6.7inch

Step-by-step explanation:

From the question, we are informed that Emma made a scale model of the Sam Houston's monument and that the monument's actual height is 67 ft tall but Emma is using a scale where 1 inch is 10 ft.

The height of Emma's model will therefore be 67ft divided by 10 since he's using a scale of 1:10. This will be:

= 67/10

= 6.7inch

6 0

3 years ago